Mole fraction
Mole fraction can be defined as the ratio of number of moles of the component in the solution to the total number of moles of all components in the solution.
Problem 1. Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
Solution.
Total mass of the solution = 100 g
Mass of benzene = 30 g.
∴ Mass of carbon tetrachloride = (100  30)g = 70 g
Molar mass of benzene (C_{6}H_{6}) = (6 × 12 + 6 × 1) g mol^{– 1} = 78 g mol^{1}
∴ Number of moles of C_{6}H_{6} =30/78 mol
= 0.3846 mol
Molar mass of carbon tetrachloride (CCl_{4}) = 1 × 12 + 4 × 355 = 154 g mol^{1}
∴ Number of moles of CCl_{4} = 70/154 mol = 0.4545 mol
Mole fraction of C_{6}H_{6}
Problem 2. A tank is charged with a mixture of 1.0 × 10^{3} mol of oxygen and 4.5 × 10^{3} mol of helium. Calculate the mole fraction of each gas in the mixture.
Solution.
The given parameters are
N_{He} = 4.5 × 10^{3} mol and N_{O}_{2} = 1.0 × 10^{3} mol
Mole fraction can be calculated as
X_{He} = 4.5 × 10^{3} mol / 4.5 × 10^{3} mol + 1.0 × 10^{3} mol
X_{He }= 4.5 mol / 5.5 mol
X_{He }= 0.82
X_{O2} = 2.0 × 10^{3} mol / 4.5 × 10^{3} mol + 1.0 × 10^{3} mol
X_{O2 }= 1.0 × 10^{3} / 5.5 × 10^{3}
X_{O2 }= 0.18
Problem 3. Determine the mole fraction of methanol CH_{3}OH and water in a solution prepared by dissolving 4.5 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.
Solution.
Moles of CH_{3}OH = 4.5 / 32 = 0.14 mole
Moles of H_{2}O = 40 / 18 = 2.2 moles
Therefore, according to the equation
mole fraction of CH_{3}OH = 0.14 / 2.2 + 0.14
= 0.061
Problem 4. What is the mole fraction of carbon tetrachloride (CCl4) in solution if 3.47 moles of CCl4 is dissolved in 8.54 moles of benzene (C_{6}H_{6})?
Solution.
Problem 5. What is the mole fraction of formaldehyde (CH_{2}O) in solution if 25.7 grams of CH_{2}O is dissolved in 3.25 moles of carbon tetrachloride (CCl_{4})?
Solution.
The grams of acetone will need to be converted into moles in order to solve for mole fraction.
Problem 6. 0.100 mole of NaCl is dissolved into 100.0 grams of pure H_{2}O. What is the mole fraction of NaCl?
Solution.
100.0 g / 18.0 g mol^{}^{1} = 5.56 mol of H_{2}O
Add that to the 0.100 mol of NaCl = 5.56 + 0.100 = 5.66 mol total
Mole fraction of NaCl = 0.100 mol / 5.66 mol = 0.018
The mole fraction of the H_{2}O:
5.56 mol / 5.66 mol = 0.982
Problem 7. A solution is prepared by mixing 25.0 g of water, H_{2}O, and 25.0 g of ethanol, C_{2}H_{5}OH. Determine the mole fractions of each substance.
Solution.
(1) Determine the moles of each substance:
H2O ⇒ 25.0 g / 18.0 g/mol = 1.34 mol
C_{2}H_{5}OH ⇒ 25.0 g / 46.07 g/mol = 0.543 mol
(2) Determine mole fractions:
H_{2}O ⇒ 1.34 mol / (1.34 mol + 0.543 mol) = 0.71
C_{2}H_{5}OH ⇒ 0.543 mol / (1.34 mol + 0.543 mol) = 0.29
Problem 8. A solution contains 10.0 g pentane, 10.0 g hexane and 10.0 g benzene. What is the mole fraction of hexane?
Solution.
(1) You need to determine the moles of pentane, hexane and benzene:
to do this, you need the molecular weights. Here are the formulas:
pentane: C_{5}H_{12}
hexane: C_{6}H_{14}
benzene: C_{6}H_{6}
(2) When you have the moles of each, add them together.
(3) Then, divide the moles of hexane by the total.
Problem 9. The molality of an aqueous solution of sugar (C_{12}H_{22}O_{11}) is 1.62m. Calculate the mole fractions of sugar and water.
Solution.
(1) Molality is moles solute / kg of solvent. Therefore we know our solution is:
1.62 mol C_{12}H_{22}O_{11}
1.00 kg = 1000 g of water
(2) Calculate the moles of water present:
1000 g / 18.0152 g/mol = 55.50868 mol
(3) Determine the mole fraction of the sugar:
1.62 mol / (1.62 mol + 55.50868 mol) = 0.028357 = 0.0284 (to three sf)
(4) You can calculate the mole fraction of the water by subtraction.
Problem 10. How many grams of water must be used to dissolve 100.0 grams of sucrose (C_{12}H_{22}O_{11}) to prepare a 0.020 mole fraction of sucrose in the solution?
Solution.
(1) Determine moles of sucrose:
100.0 g / 342.2948 g/mol = 0.292145835 mol
(2) Determine moles of water required to make the solution 0.020 mole fraction of sucrose:
0.020 = 0.292 / (0.292 + x)
(0.020) (0.292 + x) = 0.292
0.00584 + 0.02x = 0.292
0.02x = 0.28616
x = 14.308 mol of H_{2}O
Comment: you can also do this:
0.292 is to 0.02 as x is to 0.98
(3) Determine grams of water:
14.308 mol × 18.015 g/mol = 258.0 g
Problem 11. Surprisingly, water (in the form of ice) is slightly soluble in liquid nitrogen. At 196 °C, (the boiling point of liquid nitrogen) the mole fraction of water in a saturated solution is 1.00 × 10^{5}. Compute the mass of water that can dissolve in 1.00 kg of boiling liquid nitrogen.
Solution.
(1) Use the definition of mole fraction to set up the following:
χ_{water} = moles water / (moles water + moles nitrogen)
1.00 × 10^{5} = x / (x + 71.3944041)
I'm going to carry some guard digits until the end of the calculation.
(2) Some algebra:
(1.00 x 10^{5}) (x) + 7.139440411 × 10^{4} = x
0.99999x = 7.139440411 x 10^{4}
x = 7.139511806 × 10^{4} mol of H_{2}O
(3) Calculate grams of water from moles of water:
7.139511806 × 10^{4} mol × 18.0152 g/mol = 1.2862 × 10^{2 }g
1.29 × 10^{2} g (to three sf)
Problem 12. What is the mole fraction of cinnamic acid in a mixture that is 50.0% weight urea in cinnamic acid (urea = 60.06 g/mol; cinnamic acid = 148.16 g/mol)
Solution.
(1) Let us assume 100.0 g of this mixture are present.
Therefore:
50.0 g is urea
50.0 g is cinnamic
(2) Convert grams to moles:
urea: 50.0 g / 60.06 g/mol = 0.8325 mol
cinnamic acid: 50.0 g / 148.16 g/mol = 0.3375 mol
(3) Determine mole fraction of cinnamic acid:
0.3375 mol / 1.1700 mol = 0.2885
117 videos225 docs239 tests

1. What is mole fraction and how is it calculated? 
2. How does mole fraction relate to the physical properties of a solution? 
3. Can mole fraction be used to calculate the mass percentage of a component in a solution? 
4. How does the mole fraction change with temperature and pressure? 
5. Can mole fraction be used to compare the concentrations of different solutions? 

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