problem solving on mole fraction

• Science Notes Posts
• Contact Science Notes
• Todd Helmenstine Biography
• Anne Helmenstine Biography
• Free Printable Periodic Tables (PDF and PNG)
• Periodic Table Wallpapers
• Interactive Periodic Table
• Periodic Table Posters
• Science Experiments for Kids
• How to Grow Crystals
• Chemistry Projects
• Fire and Flames Projects
• Holiday Science
• Chemistry Problems With Answers
• Physics Problems
• Unit Conversion Example Problems
• Chemistry Worksheets
• Biology Worksheets
• Periodic Table Worksheets
• Physical Science Worksheets
• Science Lab Worksheets
• My Amazon Books

Mole Fraction Formula and Calculation

In chemistry, the mole fraction is a unit of concentration that is the number of moles of a component divided by the total number of moles of a solution or mixture . The mole fraction is a dimensionless number. The sum of all of the mole fractions equals 1. The symbol for mole fraction is the capital letter X or the lowercase Greek letter chi ( χ ). The terms “amount fraction” or “amount-of-substance fraction” mean the same as mole fraction.

Mole Fraction Formula

The formula for mole fraction is the moles of one component divided by the total number of moles:

X A = moles A / total moles

How to Calculate Mole Fraction

For example, in a mixture consisting of 0.25 moles of component A and 0.40 moles of component B, you can find the mole fractions of A and B.

X A = moles A / total moles = 0.25 / (0.25 + 0.40) = 0.38 (rounded)

X B = moles B / total moles = 0.40 / (0.25 + 0.40) = 0.62 (rounded)

Remember, adding up the mole fractions equals 1.

X A + X B = 1

0.38 + 0.62 = 1

If the mixture consists of more than two components, the same rules apply.

Mole Percent

A related term is mole percent. Mole percent or mole percentage is the mole fraction multiplied by 100%.

mol% = X A x 100%

The sum of all of the mole percents of a mixture equals 100%

Mole Fraction Properties and Advantages

The mole fraction offers advantages over some of the other units of concentration .

• Unlike molarity , mole fraction is not temperature dependent.
• Preparing a solution using mole fraction is easy because you simply weigh the masses of the components and then combine them.
• There is no confusion over which component is the solvent and which is the solute . The unit is symmetric in this respect because the roles of solute and solvent are reversible, depending on the mole fraction.
• In a mixture of ideal gases or most real gases, the mole fraction is the same as the ratio of partial pressure of a gas to the total pressure of the mixture. In other words, mole fraction follows Dalton’s law of partial pressure .

Example Calculations

Simple example.

For example, find the mole fraction of carbon tetrachloride in a mixture consisting of 1 mole benzene, 2 moles carbon tetrachloride, and 7 moles acetone.

X CCl4 = 2 / (1 + 2 + 7) = 2/10 = 0.2

Mole Fraction From Grams

Find the mole fraction of formaldehyde (CH 2 O) when you dissolve 25.7 grams of CH 2 O in 3.25 moles of carbon tetrachloride (CCl 4 ).

Here, the amount of CCl 4 is already in moles, but you can’t find mole fraction until you convert grams of CH 2 O into moles, too. Look up the atomic masses of carbon, hydrogen, and oxygen on the periodic table and use the chemical formula for formaldehyde to calculate the number of moles.

1 mole CH 2 O = 12.01 g + 2×1.01 g + 16.00 g = 30.03 g

Use this relationship and find the number of moles of CH 2 O.

moles CH 2 O = 25.7 g x (1 mol/30.03 g) = 0.856 mol

Now, solve for mole fraction.

X A = 0.856 moles CH 2 O / (0.856 moles CH 2 O + 3.25 moles CCl 4 ) = 0.208

How to Find Mole Fraction From Molality

Molality (m) is the moles of solute per kilogram of solvent. Using these units, you can calculate mole fraction if you know molality. For example, find the mole fraction of table sugar or sucrose (C 6 H 12 O 6 ) in a 1.62 m solution of sucrose in water.

Given the definition of molality, you know the following:

1.2 m sucrose = 1.62 moles sucrose / 1 kg water

Next, find how many moles there are of water. Use the atomic masses from the periodic table and find that the molar mass of water is 18.0 (2×1.01 + 16.00).

1 kg = 1000 g = 1 mol / 18.0 g = 55.5 moles H 2 O

Knowing the moles of sucrose and the moles of water, find the mole fraction of sucrose.

X sucrose = moles sucrose / total moles = 1.62 / (1.62 + 55.5) = 0.0284

With small numbers like this, it’s often better expressing mole fraction as mole percent. The solution is 2.84% sugar in water.

• IUPAC (1997). “Amount fraction.” Compendium of Chemical Terminology (the “Gold Book”) (2nd ed.). Blackwell Scientific Publications. ISBN 0-9678550-9-8. doi: 10.1351/goldbook.A00296
• Rickard, James N.; Spencer, George M.; Bodner, Lyman H. (2010). Chemistry: Structure and Dynamics (5th ed.). Hoboken, N.J.: Wiley. ISBN 978-0-470-58711-9.
• Thompson, A.; Taylor, B. N. (2009). “ Special Publica tion 811 .” The NIST Guide for the use of the International System of Units. National Institute of Standards and Technology.
• Zumdahl, Steven S. (2008). Chemistry (8th ed.). Cengage Learning. ISBN 978-0-547-12532-9.

Related Posts

• Chemistry Formulas

Mole Fraction Formula

What is Mole fraction?

Mole fraction represents the number of molecules of a particular component in a mixture divided by the total number of moles in the given mixture. It’s a way of expressing the concentration of a solution.

Mole Fraction formula

The molar fraction can be represented by X. If the solution consists of components A and B, then the mole fraction is,

Therefore, the sum of mole fraction of all the components is always equal to one.

Please note that mole fraction represents a fraction of molecules, and since different molecules have different masses, the mole fraction is different from the mass fraction.

Example 1  

A tank is charged with a mixture of 1.0 x 10 3 mol of oxygen and 4.5 x 10 3 mol of helium. Calculate the mole fraction of each gas in the mixture.

The given parameters are

N He = 4.5 x 10 3 mol and N O 2 = 1.0 x 10 3 mol

Mole fraction can be calculated as

X He = 4.5 x 10 3 mol / (4.5 x 10 3 mol + 1.0 x 10 3 mol)

X He =  4.5 mol / 5.5 mol

X He  = 0.82

X O 2 = (2.0 x 10 3 mol / (4.5 x 10 3 mol + 1.0 x 10 3 mol)

X O2   = 1.0 x 10 3 / 5.5 x 10 3

 X O2  = 0.18

Determine the mole fraction of CH 3 OH and H 2 O in a solution prepared by dissolving 5.5 g of alcohol in 40 g of H 2 O. M of H 2 O is 18 and M of CH 3 OH is 32.

Moles of CH 3 OH = 5.5 / 32 = 0.17 mole

Moles of H 2 O = 40 / 18 = 2.2 moles

Therefore, according to the equation

mole fraction of CH 3 OH = 0.17 / 2.2 + 0.17

mole fraction of CH 3 OH = 0.073

To solve more examples on Mole fraction formula and practice more question, please visit Byju’s.com

Leave a Comment Cancel reply

Your Mobile number and Email id will not be published. Required fields are marked *

Request OTP on Voice Call

Post My Comment

Register with BYJU'S & Download Free PDFs

Register with byju's & watch live videos.

8.010 m means 8.010 mol / 1 kg of solvent (8.010 mol) (98.0768 g/mol) = 785.6 g of solute 785.6 g + 1000 g = 1785.6 g total for solute and solvent in the 8.010 m solution. 1785.6 g / 1.354 g/mL = 1318.76 mL 8.01 moles / 1.31876 L = 6.0739 M 6.074 M (to four sig figs)

1 L of solution = 1000 mL = 1000 cm 3 1.329 g/cm 3 times 1000 cm 3 = 1329 g (the mass of the entire solution) 1329 g minus 571.4 g = 757.6 g = 0.7576 kg (the mass of water in the solution) 571.4 g / 98.0768 g/mol = 5.826 mol of H 2 SO 4 5.826 mol / 0.7576 kg = 7.690 m

mass of acetone: (3.30 mL) (0.789 g/mL) = 2.6037 g moles of acetone: 2.6037 g / 58.0794 g/mol = 0.04483 mol mass of solution: (75.0 mL) (0.993 g/mL) = 74.475 g mass of water in the solution: 74.475 g - 2.6037 g = 71.8713 g moles of water: 71.8713 g / 18.015 g/mol = 3.9896 mol

0.04483 mol / 0.0750 L = 0.598 M

0.04483 mol / 0.0718713 kg = 0.624 m

0.04483 mol / (0.04483 mol + 3.9896 mol) = 0.0111

(15.00 mol/L) (1.000 L) = 15.00 mole of HCl 15.00 mol times 36.4609 g/mol = 546.9135 g of HCl

(1000. mL) (1.0745 g/cm 3 ) = 1074.5 g of solution 1074.5 g minus 546.9135 g = 527.5865 g of water = 0.5275865 kg

15.00 mol / 0.5275865 kg = 28.43 m (to four sig figs)

0.449 mol/kg = x / 2.92 kg x = 1.31108 mol of KBr

(1.31108 mol) (119.0023 g/mol) = 156 g KBr

156 g KBr + 2920 g water = 3076 g total If you wanted to be real technical about it, then use three sig figs to obtain 3080 g.

(0.391 mol) ( 86.1766 g/mol) = 33.6950 g 33.6950 g + 1000 g = 1033.6950 g In other words, every 1033.6950 g of 0.391 m solution delivers 33.6950 g of hexane 33.6950 is to 1033.6950 as 247 is to x x = 7577.46446 g to three sig figs, 7.58 kg of solution

1.42 m means 1.42 mole of C 6 H 6 in 1 kg of tetrahydrofuran (1.42 mol) ( 78.1134 g/mol) = 110.921 g 110.921 g + 1000 g = 1110.921 g 110.921 is to 1110.921 as x is to 1630 x = 162.75 g To check, do this: 162.75 g / 78.1134 g/mol = 2.08351 mol 1630 g − 162.75 g = 1467.25 g 2.08351 mol / 1.46725 kg = 1.42 m

A mole fraction of 0.100 for NaCl means the mole fraction of water is 0.900. Let us assume a solution is present made up of 0.100 mole of NaCl and 0.900 mole of water. mass of water present ---> (0.900 mol) (18.015 g/mol) = 16.2135 g molality of solution ---> 0.100 mol / 0.0162135 kg = 6.1677 m to three sig figs, 6.17 m

mass solvent ---> 7550 g − 929 g = 6621 g = 6.621 kg moles solute ---> 929 g/ 84.93 g/mol = 10.9384 mol molality = 10.9384 mol / 6.621 kg = 1.65 m

(1000 mL) (1.230 g/mL) = 1230 g

(3.75 mol) (98.0768 g/mol) = 367.788 g

1230 − 367.788 = 862.212 g

3.75 mol / 0.862212 kg = 4.35 molal (to three sig figs)

1.00 L of this solution contains 4.20 mole of NaCl. (1.00 L) (1050 g/L) = 1050 g of solution.

(4.20 mol) (58.443 g/mol) = 245.4606 g 1050 g - 245.4606 g = 804.5394 g

4.20 mol / 0.8045394 kg = 5.22 m (to three sig figs)

3.58 mole of RbCl in 1000 g of water.

(3.58 mol) (120.921 g/mol) = 432.89718 g 1000 g + 432.89718 g = 1432.89718 g

1432.89718 g / 1.12 g/mL = 1279.37 mL

3.58 mol / 1.27937 L = 2.80 M

molality = moles of naphthalene / kilograms of benzene (16.5 g / 128.1732 g/mol) / 0.0543 kg = 2.37 m

(1.34 mL) (1.59 g/mL) = 2.1306 g 2.1306 g / 153.823 g/mol = 0.013851 mol

(65.0 mL) (1.33 g/mL) = 86.45 g = 0.08645 kg

0.013851 mol / 0.08645 kg = 0.160 m (to three sig figs)

MV = mass / molar mass (x) (0.4500 L) = 0.825 g / 141.9579 g/mol x = 0.0129 M

0.825 g / 141.9579 g/mol = 0.00581158 mol 0.00581158 mol / 0.4500 kg = 0.0129 m

Na 2 HPO 4 ---> 0.825 g / 141.9579 g/mol = 0.00581158 mol H 2 O ---> 450.0 g / 18.015 g/mol = 24.97918401 mol mole fraction of the water ---> 24.97918401 mol / (24.97918401 mol + 0.00581158 mol) = 0.9998

water ---> (450 g / 450.825 g) (100) = 99.8%

ppm means the number of grams of solute per 1,000,000 grams of solution 0.825 is to 450.825 as x is to 1,000,000 x = 1830 ppm

    mol solute m  =  –––––––––     kg solvent

    mass solute n  =  –––––––––––––––––     molar mass of solute

I'm going to use the kg/mol amount and the reason will show up in a moment.     mass solute n  =  ––––––––––––     0.0580794 kg/mol

mass of solution = mass of solvent + mass of solute = 450.0 g = 0.4500 kg I'll ignore those two trailing zeros for the moment. We write this: mass of solvent = 0.45 − mass of solute Here's the reason why I must use the kg/mol unit: In the subtraction just above, the 0.45 is in kilograms. That means the mass of solute must also be in kilograms. You can't subtract two numbers using different units. Also, the bottom unit must be in kilograms because the 0.75 molal value is determined with kg in the denominator. Using grams in the denominator is not done with molality.

  x –––––– 0.0580794 0.75 = ––––––––––––   0.45 − x     x 0.3375 − 0.75x  =  –––––––––     0.0580794 x = 0.019602 − 0.04356x 1.04356x = 0.019602 x = 0.01878 kg = 18.78 g (to 4 sig figs)

0.7500 molal means 0.7500 mole of solute (the acetone) per 1000 g of water mass of acetone ---> 58.0794 g/mol times 0.7500 mol = 43.56 g mass of solution ---> 1000 g + 43.56 g = 1043.56 g 43.56 is to 1043.56 as x is to 450 x = 18.78 g

Numerical Problems on Molality

• Post author By Hemant More
• Post date January 30, 2020
• 23 Comments on Numerical Problems on Molality

Science > Chemistry > Solutions and Their Colligative Properties > Numerical Problems on Molality

In this article, we shall study numerical problems to calculate molality of a solution.

Example – 01:

7.45 g of potassium chloride (KCl) was dissolved in 100 g of water. Calculate the molality of the solution.

Given: mass of solute (KCl) = 7.45 g, mass of solvent (water) = 100 g = 0.1 kg

To Find: Molarity of solution =?

Molecular mass of KCl = 39 g x 1 + 35.5 g x 1 = 74.5 g mol -1

Number of moles of solute (KCl) = given mass/ molecular mass

Number of moles of solute (KCl) = 7.45 g/ 74.5 g mol -1 = 0.1 mol

Molality = Number of moles of solute/Mass of solvent in kg

Molality = 0.1 mol /0.1 kg = 1 mol kg -1

Ans: The molality of solution is 1 mol kg -1  or 1 m.

Example – 02:

11.11 g of urea (NH 2 CONH 2 ) was dissolved in 100 g of water. Calculate the molarity and molality of the solution. Given N = 14, H = 1, C = 12, O = 16.

Given: mass of solute (urea) = 11.11 g, mass of solvent (water) = 100 g = 0.1 kg

Molecular mass of urea (NH 2 CONH 2 ) = 14 g x 2 + 1 g x 4 + 12 g x 1 + 16 g x 1

Molecular mass of urea (NH 2 CONH 2 ) = 60 g mol -1

Number of moles of solute (urea) = given mass/ molecular mass

Number of moles of solute (urea) = 11.11 g/ 60 g mol -1 = 0.1852 mol

Volume of water = mass of water/ density = 100 g/1 g mL -1 = 100 mL = 0.1 L

Molarity = Number of moles of solute/Volume of solution in L

Molarity = 0.1852 mol /0.1 L = 1.852 mol L -1 or 1.852 mol dm -3

Molality = 0.1852 mol /0.1 kg = 1.852 mol kg -1

Ans: The molarity of solution is 1.852 mol L -1 and the molality is 1.852 mol kg -1

Example – 03:

34.2 g of sugar was dissolved in water to produce 214.2 g of sugar syrup. Calculate molality and mole fraction of sugar in the syrup. Given C = 12, H = 1 and O = 16.

Given: Mass of solute (sugar) = 34.2 g, Mass of solution (sugar syrup) = 214.2 g

To Find: Molality and mole fraction =?

Mass of Solution = Mass of solute + mass of solvent

Mass of solvent = mass of solution – mass of solute = 214.2 g – 34.2 g = 180 g = 0.180 kg

Molar mass of sugar (C 12 H 22 O 11 ) = 12 g x 12 + 1 g x 22 + 16 g x 11 = 342 g mol -1

Number of moles of solute (sugar) = n B  = Given mass/ molecular mass = 34.2 g/342 g mol -1   = 0.1 mol

Molality = 0.1 mol /0.180 kg = 0.5556 mol kg -1

Molar mass of water (H 2 O) = 1 g x 2 + 16 g x 1 = 18 g mol -1

Number of moles of solvent (water) = n A  = Given mass/ molecular mass = 180 g/18 g mol -1   = 10 mol

Total number of moles = n A  + n B  = 0.1 + 10 = 10.1 mol

Mole fraction of solute (sugarl) = x B = n B /(n A  + n B ) = 0.1/10.1 = 0.0099

Mole fraction of sugar = 0.0099

Ans: Molality of solution = 0.5556 mol kg -1  and mole fraction of sugar = 0.0099

Example – 04:

10.0 g KCl is dissolved in 1000 g of water. If the density of the solution is 0.997 g cm -3 , calculate a) molarity and b) molality of the solution. Atomic masses K = 39 g mol -1 , Cl = 35.5 g mol -1 .

Given: the mass of solute (KCl) = 10 g, the mass of solvent (water) = 1000 g = 1 kg, density of solution = 0.997 g cm -3 ,

To Find: molarity =? molality = ?

Number of moles of solute (KCl) = 10 g/ 74.5 g mol -1 = 0.1342 mol

Molality = 0.1342 mol /1 kg = 0.1342 mol kg -1

Mass of solution = 10 g + 1000 g = 1010 g

Volume of solution = mass of solution/density = 1010/0.997 g cm -3

Volume of solution = 1013 cm 3 = 1013 mL = 1.013 L

Molarity = 0.1342 mol /1.013 L = 0.1325 mol L -1

Ans: The molarity of the solution is 0.1325 mol L -1  or 0.1325 M, the molality of the solution is 0.1342 mol kg -1  or 0.1342 m.

Example – 05:

Calculate molarity and molality of the sulphuric acid solution of density 1.198 g cm -3  containing 27 % by mass of sulphuric acid.

Given: density of the solution = 1.198 g cm -3 , % mass of sulphuric acid = 27%,

To Find: Molarity =? and molality =?

Consider 100 g of solution

Mass of H 2 SO 4  = 27 g and mass of H 2 O = 100 – 27 g = 73 g = 0.073 kg

Molecular mass H 2 SO 4  = 1 g x 2 + 32 g x 1 + 16g  x 4 = 98 g mol -1

Number of moles of H 2 SO 4 = n B = 27 g/ 98 g = 0.2755 mol

Density of solution = 1.198 g cm -3

Volume of solution = Mass of solution / density = 100 g /1.198 g cm -3 = 83.47 cm 3 = 83.47 mL = 0.08347 L

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.2755/0.08347 = 3.301 M

Molality = Number of moles of solute/mass of sovent in kg

Molality = 0.2755 mol /0.073 kg = 3.774 mol L -1

Ans: The molarity of solution is 3.374 mol L -1  or 3.374 M, the molality of solution is 3.774 mol L -1  or 3.774 m

Example – 06:

Calculate the mole fraction, molality and molarity of HNO 3 in a solution containing 12.2 % HNO 3 . Given density of HNO 3 as 1.038 g cm -3 , H = 1, N = 14, O = 16.

Given: density of the solution = 1.038 g cm -3 , % mass of HNO 3  = 12.2 %,

To Find:  mole fraction =? molarity =? and molality =?

Mass of HNO 3 = 12.2 g and mass of H 2 O = 100 – 12.2 g = 87.8 g = 0.0878 kg

Molecular mass of water (H 2 O) = 1 g x 2 + 16 g x 1 = 18 g mol -1

Molecular mass HNO 3 = 1 g x 1 + 14 g x 1 + 16g  x 3 = 63 g mol -1

Number of moles of water = n A = 87.8 g/ 18 g = 4.8778 mol

Number of moles of HNO 3 = n B = 12.2 g/ 63 g = 0.1937 mol

Total number of moles = n A + n B + n C = 4.8778 + 0.1937 = 5.0715

Mole fraction of HNO 3 = x B = n B /(n A  +n B ) = 0.1937/5.0715 = 0.0382

Density of solution = 1.038 g cm -3

Volume of solution = Mass of solution / density = 100 g /1.038 g cm -3 = 96.34 cm 3 = 96.34 mL = 0.09634 L

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.1937/0.09634 =2.011 M

Molality = 0.1937 mol /0.0878 kg = 2.206 mol kg -1

Ans:  The mole fraction of HNO3 is 0. 0382, the molarity of solution is 2.011 mol L -1  or 2.011 M, the molality of solution is 2.206 mol kg -1  or 2.206 m

Example – 07:

Calculate molarity and molality of 6.3 % solution of nitric acid having density 1.04 g cm -3 . Given atomic masses H = 1, N = 14 and O = 16.

Given: density of the solution = 1.04 g cm -3 , % mass of HNO 3  = 6.3 %,

Mass of HNO 3 = 6.3 g and mass of H 2 O = 100 – 6.3 g = 93.7 g = 0.0937 kg

Number of moles of water = n A = 93.4 g/ 18 g = 5.189 mol

Number of moles of HNO 3 = n B = 6.3 g/ 63 g = 0.1 mol

Density of solution = 1.04 g cm -3

Volume of solution = Mass of solution / density = 100 g /1.04 g cm -3 = 96.15 cm 3 = 96.15 mL = 0.09615 L

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.1/0.09615 =1.040 M

Molality = 0.1 mol /0.0937 kg = 1.067 mol kg -1

Ans:  The molarity of solution is 1.040 mol L -1  or 1.040 M

The molality of solution is 1.067 mol kg -1  or 1.067 m

Example – 08:

An aqueous solution of NaOH is marked 10% (w/w). The density of the solution is 1.070 g cm -3 . Calculate molarity, molality and mole fraction of NaOH in water. Given Na = 23, H =1 , O = 16

Mass of NaOH = 10 g and mass of H 2 O = 100 – 10 g = 90 g = 0.090 kg

Molecular mass NaOH = 23 g x 1 + 16 g x 1 + 1 g  x 1 = 40 g mol -1

Number of moles of water = n A = 90 g/ 18 g = 5 mol

Number of moles of NaOH = n B = 10 g/ 40 g = 0.25 mol

Total number of moles = n A + n B = 5 + 0.25 = 5.25 mol

Mole fraction of NaOH = x B = n B /(n A  +n B ) = 0.25/5.25 = 0.0476

Density of solution = 1.070 g cm -3

Volume of solution = Mass of solution / density = 100 g /1.070 g cm -3 = 93.46 cm 3 = 93.46 mL = 0.09346 L

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.25/0.09346 =2.675 M

Molality = 0.25 mol /0.090 kg = 2.778 mol kg -1

Ans:  The molarity of solution is 2.675mol L -1  or 2.675 M, the molality of solution is 2.778 mol kg -1  or 2.778 m, the mole fraction of NaOH is 0. 0476

Example – 09:

A solution of glucose in water is labelled as 10 % (w/w). Calculate a) molality and b) molarity of the solution. Given the density of the solution is 1.20 g mL -1  and molar mass of glucose is 180 g mol -1 .

Given: density of the solution = 1.20 g cm -3 , % mass of glucose = 10 %, molar mass of glucose is 180 g mol -1 .

To Find:  molarity =? and molality =?

Mass of glucose = 10 g and mass of H 2 O = 100 – 10 g = 90 g = 0.090 kg

Molecular mass glucose = 180 g mol -1

Number of moles of glucose = n B = 10 g/ 180 g = 0.0556 mol

Density of solution = 1.20 g cm -3

Volume of solution = Mass of solution / density = 100 g /1.20 g cm -3 = 83.33 cm 3 = 83.33 mL = 0.08333 L

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.0556/0.08333 =0.6672 M

Molality = 0.0556 mol /0.090 kg = 0.6178 mol kg -1

Ans:  The molarity of solution is 0.6672 mol L -1  or 0.6672 M, the molality of solution is 0.6178 mol kg -1  or 0.6178 m,

Example – 10:

Battery acid 4.22 M aqueous H 2 SO 4 solution, and has density 1.21 g cm -3 . What is the molality of H 2 SO 4 . Given H = 1, S = 32, O = 16

Given: density of the solution = 1.21 g cm -3 , Molarity of solution = 4.22 M.

To Find:  molality =?

Let us consider 1 L of solution

Molarity of solution = Number of moles of the solute/volume of solution in L

Number of moles of solute = Molarity of solution x volume of solution in L = 4.22 x 1 = 4.22

Density of solution = 1.21 g cm -3  = 1.21 g/mL = 1.21 x 10 3  g/L = 1.21 kg/L

Mass of solution = Volume of solution x density = 1 L x 1.21 kg/L = 1.21 kg

Mass of solute (H 2 SO 4 ) = Number of moles x molecular mass = 4.22 x 98

Mass of solute (H 2 SO 4 ) = 413.56 g = 0.41356 kg

Mass of solvent = mass of solution – mass of solute = 1.21 – 0.41356 = 0.79644 kg

Molality = 4.22 mol /0.79644 kg = 5.298 mol kg -1

Ans: Molality of solution is 5.298 mol kg -1  or 5.298 m

Example – 11:

The density of 5.35 M H 2 SO 4 solution is 1.22 g cm -3 . What is molality of a solution?

Given: density of the solution = 1.22 g cm -3 , Molarity of solution = 5.35 M.

Molecular mass H 2 SO 4  = 1 g x 2 + 32 g x 1 + 16g x 4 = 98 g mol -1

Number of moles of solute = Molarity of solution x volume of solution in L = 5.35 x 1 = 5.35

Density of solution = 1.22 g cm -3  = 1.22 g/mL = 1.22 x 10 3  g/L = 1.22 kg/L

Mass of solution = Volume of solution x density = 1 L x 1.22 kg/L = 1.22 kg

Mass of solute (H 2 SO 4 ) = Number of moles x molecular mass = 5.35 x 98

Mass of solute (H 2 SO 4 ) = 524.3 g = 0.5243 kg

Mass of solvent = mass of solution – mass of solute = 1.22 – 0.5243 = 0.6957 kg

Molality = 5.35 mol /0.6957 kg = 7.690 mol kg -1

Ans: Molality of solution is 7.690 mol kg -1  or 7.690 m

Example – 12:

Calculate the mole fraction of solute in its 2 molal aqueous solution.

Given:  molality = 2 molal

To Find:  Mole fraction =?

Molality of solution = 2 molal = 2 mol mol kg -1

The number of moles of solute = 2

The mass of solvent (water) = 1 kg = 1000 g

Number of moles of solvent (water) = 1000/16 = 55.55

Mole fraction of solute = 2/(2 + 55.55) = 2/57.55 = 0.03475

Ans: Mole fraction of solute is 0.0345

Previous Topic: Numerical Problems on Molarity

Next Topic: Short Cut Methods to Calculate Concentration of Solution

• Tags Alloys , Amalgams , Aqueous solution , Chemistry , Colloidal solution , Concentration , Corse solution , Dissolving , Formality , Gaseous solutions , Grams per litre , Heterogeneous solution , Homogeneous solution , Immiscible liquids , Insoluble substance , Liquid solutions , Mass percentage , Miscible liquids , Molality , Molar concentration , Molarity , Molarity of dilution , Molarity of mixing , Mole , Mole fraction , Normality , Percentage by mass , Percentage by mass by volume , Percentage by volume , ppm , Saturated solution , Solid solutions , Solubility , Solubility curves , Soluble substance , Solute , Solution , Solvent , Strength , Supersaturated solution , Suspension , True solution , Types of solutions , Unsaturated solution Particles per million

23 replies on “Numerical Problems on Molality”

Where you found this type of questions ? Pls reply me

Past State Boards, CBSE, JEE and NEET Papers

thanks this is very helpful

Wooww, I love these questions They upgrade your thinking ability

Nice question

Really its very helpful for us thank you so much

These questions help me but I need questions of mole fraction

Problems on Mole Fraction https://thefactfactor.com/facts/pure_science/chemistry/physical-chemistry/mole-fraction/7855/ More Topics in Colligative Properties: https://thefactfactor.com/chemistry/solutions-and-their-colligative-properties/ More Topics in Chemistry: https://thefactfactor.com/chemistry/

Love this question Best thinking upgrade questions

Nice questions really help me alot 👍👍👍👌👌

Its a nice questions and not to hard

Thanx for questions……😊👍

These questions help to improve myself 😇😇 so thnx for this question 🤗🤗🤗🤗🤗

Thank you for these questions. This was very helpful ☺️

Thanks a lot.This help me so much

Thank u so much

Thanks a lot, the content is so helpful

Very bralint questions

I love you 😘😘😘

Thankyou so much

Thank you so much. Very helpful.

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

This website will show the principles of solving Math problems in Arithmetic, Algebra, Plane Geometry, Solid Geometry, Analytic Geometry, Trigonometry, Differential Calculus, Integral Calculus, Statistics, Differential Equations, Physics, Mechanics, Strength of Materials, and Chemical Engineering Math that we are using anywhere in everyday life. This website is also about the derivation of common formulas and equations. (Founded on September 28, 2012 in Newark, California, USA)

Wednesday, September 3, 2014

Mole fraction problems.

The Tech Edvocate

• Advertisement
• Home Page Five (No Sidebar)
• Home Page Four
• Home Page Three
• Home Page Two
• Icons [No Sidebar]
• Left Sidbear Page
• Lynch Educational Consulting
• My Speaking Page
• Newsletter Sign Up Confirmation
• Newsletter Unsubscription
• Page Example
• Privacy Policy
• Protected Content
• Request a Product Review
• Shortcodes Examples
• Terms and Conditions
• The Edvocate
• The Tech Edvocate Product Guide
• Write For Us
• Dr. Lynch’s Personal Website
• The Edvocate Podcast
• Assistive Technology
• Child Development Tech
• Early Childhood & K-12 EdTech
• EdTech Futures
• EdTech News
• EdTech Policy & Reform
• EdTech Startups & Businesses
• Higher Education EdTech
• Online Learning & eLearning
• Parent & Family Tech
• Personalized Learning
• Product Reviews
• Tech Edvocate Awards
• School Ratings

Hostages Killed in Gaza, Five From Supernova Festival, Spark Israel Protests

Enhancing the memory of learners: everything you need to know, iss astronaut’s stunning time-lapse includes the milky way, kamala harris says trump ‘disrespected sacred ground’ just for a ‘political stunt’ at arlington national cemetery, timeseries indexing at scale, daily horoscope: september 2, 2024, how to watch the nfl online for free in the uk, how to unblock xhamster for free, price drop: get 1tb of cloud storage for life for just £53 in september, home learning schedules for young children: everything you need to know, how to calculate a mole fraction: a step-by-step guide.

Mole fraction is a fundamental concept in chemistry that plays a crucial role in various aspects of the subject, from measuring concentrations to determining the properties of solutions. In simple terms, mole fraction refers to the ratio of the moles of one component to the total moles of all components in a mixture or solution. In this article, we will discuss step-by-step how to calculate mole fraction.

Step 1: Understanding the Formula

To calculate mole fraction, use this formula:

Mole Fraction (X) = (Moles of Component A) / (Total Moles of Components in the Mixture)

The main objective is to find the moles of each component and then divide the moles of the desired component by the total moles in the system.

Step 2: Determining Moles from Grams

If you are given mass values (in grams) for each component in a mixture and need to convert them to moles, use the following formula:

Moles = Mass (grams) / Molar Mass (g/mol)

Remember that molar mass is equal to the sum of all atomic masses multiplied by their respective ratios present in one molecule or unit cell of that substance.

Step 3: Calculating Total Moles

Now that you have converted all component masses into moles, add those mole values to find out the total number of moles present. Ensure that you have accounted for every component present in your mixture or solution.

Total Moles = Moles of Component A + Moles of Component B + …

Step 4: Finding Mole Fraction

Utilizing the obtained values, calculate the mole fraction for each component using the formula shared previously. Divide each component’s moles individually by total moles to find their corresponding mole fractions.

Mole Fraction (XComponentA) = Moles of Component A / Total Moles

Mole Fraction (XComponentB) = Moles of Component B / Total Moles

After you finish calculating the mole fractions, make sure to check if all the mole fractions add up to 1 (or nearly 1). If they do, it indicates that you have successfully computed your mole fractions.

Example Problem:

We have a solution containing 10 grams of glucose (C6H12O6), 20 grams of fructose (C6H12O6), and 50 grams of water (H2O). Calculate the mole fraction for each component.

Step 1: Convert mass to moles.

Moles of Glucose: 10 g / 180.16 g/mol ≈ 0.0555 mol

Moles of Fructose: 20 g / 180.16 g/mol ≈ 0.1111 mol

Moles of Water: 50 g / 18.0152 g/mol ≈ 2.7770 mol

Step 2: Calculate total moles.

Total Moles: ≈ 0.0555 + 0.1111 + 2.7770 ≈ 2.9435 mol

Step 3: Find the mole fraction for each component.

Mole Fraction (Glucose): ≈ 0.0555 / 2.9435 ≈ 0.0189

Mole Fraction (Fructose): ≈ .1111 / 2.9435 ≈ .0377

Mole Fraction (Water): ≈ .7770/2.9435 ≈ .9434

Now you know how to calculate mole fractions based on given data for any mixture or solution, allowing you to solve problems in chemistry more effectively and accurately!

4 Ways to Make a Veil

5 ways to help a cat cough ....

Matthew Lynch

Related articles more from author.

How to calculate mortgage payoff when selling home

How to calculate relative humidity

How to calculate of a number

How to calculate molar mass of a compound

How much usda loan do i qualify for calculator

How is Alimony Calculated in Pennsylvania

Stack Exchange Network

Stack Exchange network consists of 183 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

Q&A for work

Connect and share knowledge within a single location that is structured and easy to search.

Finding mole fraction from molality

Calculate the mole fraction of ammonia in a $\pu{2.00 m}$ solution of $\ce{NH3}$ in water.

What I know is that the formula for mole fraction is

$$X = \frac{\text{no.-of-moles-of-solute}}{\text{(no.-of-moles-of-solute)} + \text{(no.-of-moles-of-solvent)}}$$

The solute is ammonia which is $\ce{NH3}$ with a molar mass (MM) of $\pu{17 g mol-1}$ , whereas the solvent is water or $\ce{H2O}$ which has a molar mass of $\pu{18 g mol-1}$ .

The $\pu{2.00 m}$ from the problem signifies molality (because of the small $\pu{m}$ ), and molality is

$$\frac{\text{no.-of-moles-of-solute}}{\text{mass-of-solvent-in-kg}}$$

With no. of moles

$$n = \frac{m}{\text{MM}}$$

Despite knowing the formulas, I can't seem to solve for the answer. The answer should be $0.0347$ , but I can't seem to get the right solution.

Any help would be appreciated.

• aqueous-solution
• stoichiometry
• concentration

• 2 $\begingroup$ Please note: 1. The quantity ‘amount of substance’ shall not be called ‘number of moles’, just as the quantity ‘mass’ shall not be called ‘number of kilograms’. 2. Descriptive terms or names of quantities shall not be arranged in the form of an equation. 3. Multiletter abbreviated terms (such as ‘MM’) shall not be used in the place of symbols. $\endgroup$ –  user7951 Commented Jan 20, 2019 at 17:40

2 Answers 2

Despite the unconventional notations, your formula is generally correct; however, you should've express mole fraction via molality explicitly and only then plug in the numbers. By definition, mole fraction of $i$ -th component $x_i$ is

$$x_i = \frac{n_i}{n_\mathrm{tot}}$$

where $n_i$ – amount of $i$ -th component; $n_i$ – total amount of all mixture constituents. For a simple solution of a single component the following hold true:

$$x_i = \frac{n_i}{n_i + n_\mathrm{solv}}$$

where $n_\mathrm{solv}$ – amount of the solvent that can also be found via its molecular mass $M_\mathrm{solv}$ and mass $m_\mathrm{solv}$ , which, in turn, appears in the expression for molarity $b_i$ :

$$b_i = \frac{n_i}{m_\mathrm{solv}} \quad\implies\quad m_\mathrm{solv} = \frac{n_i}{b_i}$$

$$n_\mathrm{solv} = \frac{m_\mathrm{solv}}{M_\mathrm{solv}} = \frac{n_i}{b_iM_\mathrm{solv}}$$

Finally, mole fraction can be expressed via molality as follows:

$$ \require{cancel} x_i = \frac{n_i}{n_i + n_\mathrm{solv}} = \frac{n_i}{n_i + \frac{n_i}{b_iM_\mathrm{solv}}} = \frac{\cancel{n_i}}{\cancel{n_i}\left(1 + (b_iM_\mathrm{solv})^{-1}\right)} = \frac{1}{1 + (b_iM_\mathrm{solv})^{-1}} $$

Time to plug in the numbers:

$$ \begin{align} x_i &= \frac{1}{1 + (b_iM_\mathrm{solv})^{-1}}\\ &= \frac{1}{1 + (\pu{2.00e-3 mol g-1}\cdot\pu{18.02 g mol-1})^{-1}}\\ &\approx 0.0347 \end{align} $$

Few key points:

• Note that you have to convert molality expressed in $\pu{mol \color{red}{kg}-1}$ before plugging in the value: $$\pu{1 m} = \pu{1 mol kg-1} = \pu{1e-3 mol g-1}$$
• In general, never omit units in your calculations and use standardized notations.
• Mind significant figures. Since molality is given with two decimal points, you also should've taken molecular mass with higher precision.
• $\begingroup$ Thank you. I would like to ask some questions though. 1. Xi stands for mole fraction of i-th component, so if for example I was asked to find the mole fraction of the solvent, instead of the solute, will the formula be the same? 2. The reason for expressing the molality in mol kg^-1 is so that it will have the same unit as the molar mass of the solvent? 3. This is too much to ask but can you answer the problem by using the formulas I've written above (if it's possible). Or at least how to transform / derive it into your kinda shortcut formula. Again, thank you~ $\endgroup$ –  Jayce Commented Jan 20, 2019 at 13:21
• $\begingroup$ 1. Yes, with respect to molar mass of solute, or just use $x_\mathrm{solv}=1-x_i$ for a single dissolved component; 2. No, 1 molal solution is a solution of 1 mol of the given compound in 1 kg of solvent by definition (not related to molar mass at all); 3. Since you used non-standard notations (or none at all) I'd rather prefer not to do that as it's going to bring a lot of confusion on both sides; I'll try to post an updated answer with the derivation later this day. $\endgroup$ –  andselisk ♦ Commented Jan 20, 2019 at 13:28
• $\begingroup$ @Jayce The answer is updated with the derivation of the formula linking molality with mole fraction $\endgroup$ –  andselisk ♦ Commented Jan 20, 2019 at 13:53
• $\begingroup$ Thank you again. It is clear now how the formula was derived. One of the reason I got too confused in answering the problem was due to the line in question: "2.00m solution of NH3". I assumed that the 2 molal is the molality of ammonia and not the solvent / water. Another reason was I kept on figuring out how can I insert the molar mass of NH3 into the formula and also how can I find the mass of water and ammonia given the limited givens. Again, thank you. I learned a new formula, thanks to you~ $\endgroup$ –  Jayce Commented Jan 20, 2019 at 14:49
• $\begingroup$ @Jayce No prob, and good luck with chemistry:) $\endgroup$ –  andselisk ♦ Commented Jan 20, 2019 at 14:49

You don't have to memorize some weird formula like andselisk has proposed.

You have sufficient information to solve the problem:

Calculate the mole fraction of ammonia in a $\pu{2.00 molal}$ solution of $\ce{NH3}$ in water.

We can assume any quantity of solution, so let's assume 1.00 kg of solvent. So the mass of solvent (water) is $\pu{1 kilogram} = \pu{1000 g}$ in a molal solution by definition.

moles of water = $\dfrac{1000}{18.015} = 55.402$

For 1.00 kg of solvent there are 2 moles of $\ce{NH3}$ which has a mass of $\pu{2 moles}\times \pu{17.031 g/mole} = \pu{34.062 g}$

From the Op's formula:

$X = \frac{\text{no.-of-moles-of-solute}}{\text{(no.-of-moles-of-solute)} + \text{(no.-of-moles-of-solvent)}} = \dfrac{2}{2 + 55.402} \approx 0.0348$

Now I'll confess that the significant figures in this problem bother me. To have three significant figures the molality should have been given as 2.00 molal, not 2 molal.

• $\begingroup$ Thank you. To be honest, I avoid memorizing too much formula. What confuses me though (until now) is the line "2.00m solution of NH3 in water". How did you know that there are 2 "moles" of NH3? Since that "2" from the question is the molal solution or molality of ammonia = 2 and its unit is mol/kg which is not the same with the number of moles (n), which is just "mol". Sorry for such question, I'm new to this. $\endgroup$ –  Jayce Commented Jan 21, 2019 at 5:07
• $\begingroup$ @Jayce - The problem is open ended, so one can assume as much solution as desired. Frankly I tried solving the problem as 2 molar ( ie 1 liter of solution) which gave the "wrong" answer. Then I tried 2 molal (ie 1 kg of solvent) and got the "right" answer. An old convention is to use M for molar and m for molal. But without knowing which convention the particular book is using, that is somewhat of a guess. I think the newer convention is to be more explicit and use mol/L and mol/kg . $\endgroup$ –  MaxW Commented Jan 21, 2019 at 7:41
• $\begingroup$ @Jayce - I edited the solution and moved things around a bit. Does that make the line of thought clearer? $\endgroup$ –  MaxW Commented Jan 21, 2019 at 8:33

Your Answer

Reminder: Answers generated by artificial intelligence tools are not allowed on Chemistry Stack Exchange. Learn more

Sign up or log in

Post as a guest.

Required, but never shown

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy .

Not the answer you're looking for? Browse other questions tagged aqueous-solution stoichiometry concentration or ask your own question .

• Featured on Meta
• Bringing clarity to status tag usage on meta sites
• Announcing a change to the data-dump process

Hot Network Questions

• Using ON-ON switch instead of ON-OFF switch?
• What should I do if my student has quarrel with my collaborator
• Maximizing the common value of both sides of an equation (part 2)
• Citrix published app asks for credentials
• Do all instances of a given string get replaced under a rewrite rule?
• Light switch that is flush or recessed (next to fridge door)
• In what instances are 3-D charts appropriate?
• Has any astronomer ever observed that after a specific star going supernova it became a Black Hole?
• A novel (and a movie) about a village in southern France where a strange succession of events happens
• Geometry optimization of two or more different interacting molecules
• Is it a date format of YYMMDD, MMDDYY, and/or DDMMYY?
• If I am to use midi keyboard only, do I still need audio interface?
• Is it possible to travel to USA with legal cannabis?
• Strange variable scope behavior when calling function recursivly
• How would humans actually colonize mars?
• TikZ: Draw Boxes in Field of Numbers
• Fetch grapghQl response in next js
• Why is there so much salt in cheese?
• Uniqueness of Neumann series
• Is it safe to install programs other than with a distro's package manager?
• Why are poverty definitions not based off a person's access to necessities rather than a fixed number?
• Can Christian Saudi Nationals visit Mecca?
• How to translate the German word "Mitmenschlich(keit)"
• How to add a third argument to a popup_menu() callback function?

What Is a Mole Fraction?

• Chemical Laws
• Periodic Table
• Projects & Experiments
• Scientific Method
• Biochemistry
• Physical Chemistry
• Medical Chemistry
• Chemistry In Everyday Life
• Famous Chemists
• Activities for Kids
• Abbreviations & Acronyms
• Weather & Climate
• Ph.D., Biomedical Sciences, University of Tennessee at Knoxville
• B.A., Physics and Mathematics, Hastings College

Mole fraction is a unit of concentration , defined to be equal to the number of moles of a component divided by the total number of moles of a solution. Because it is a ratio, mole fraction is a unitless expression. The mole fraction of all components of a solution, when added together, will equal 1.

Mole Fraction Example

In a solution of 1 mol benzene, 2 mol carbon tetrachloride, and 7 mol acetone, the mole fraction of the acetone is 0.7. This is determined by adding up the number of moles of acetone in the solution and dividing the value by the total number of moles of components of the solution:

Number of Moles of Acetone: 7 moles

Total Number of Moles in Solution = 1 moles (benzene) + 2 moles (carbon tetrachloride) + 7 moles (acetone) Total Number of Moles in Solution = 10 moles

Mole Fraction of Acetone = moles acetone / total moles solution Mole Fraction of Acetone = 7/10 Mole Fraction of Acetone = 0.7

Similarly, the mole fraction of benzene would be 1/10 or 0.1 and the mole fraction of carbon tetrachloride would be 2/10 or 0.2.

• What Is a Mole in Chemistry?
• Mole Definition in Chemistry
• What Is Chemistry? Definition and Description
• Chemistry Vocabulary Terms You Should Know
• Why Is a Mole in Chemistry Called a Mole?
• The 7 Base Units of the Metric System
• Heat of Formation Definition - Chemistry Glossary
• Molecular Formula Definition
• Molecular Mass Definition
• Acid Anhydride Definition in Chemistry
• Chemistry 101 - Introduction & Index of Topics
• Activation Energy Definition in Chemistry
• Hess's Law Definition
• Covalent or Molecular Compound Properties
• Allotrope Definition and Examples
• The Difference Between Molality and Molarity