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**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.

- It is denoted by the alphabet x and subscript written on the right hand side of x denotes the component of which mole fraction is being calculated.
- Mathematically, Mole fraction of a component
- Let us consider a binary mixture of A and B. Let the number of moles of A and B bee n
_{A}and n_{B}respectively, then - For solution where number of components = i
- In a given solution sum of the mole fractions of all the components is unity.

Mathematically, x_{1}+ x_{2}+ x_{3}….+ x_{i}= 1

**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. **

Moles of CH

Moles of H

Therefore, according to the equation

mole fraction of CH

= 0.061

**Problem 5. **

The grams of acetone will need to be converted into moles in order to solve for mole fraction.

**Problem 6. **

100.0 g / 18.0 g mol

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

5.56 mol / 5.66 mol = 0.982

H2O ⇒ 25.0 g / 18.0 g/mol = 1.34 mol

C

H

C

to do this, you need the molecular weights. Here are the formulas:

pentane: C

hexane: C

benzene: C

Solution.

1.62 mol C

1.00 kg = 1000 g of water

1000 g / 18.0152 g/mol = 55.50868 mol

1.62 mol / (1.62 mol + 55.50868 mol) = 0.028357 = 0.0284 (to three sf)

100.0 g / 342.2948 g/mol = 0.292145835 mol

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

Comment: you can also do this:

0.292 is to 0.02 as x is to 0.98

14.308 mol × 18.015 g/mol = 258.0 g

χ

1.00 × 10

I'm going to carry some guard digits until the end of the calculation.

(1.00 x 10

0.99999x = 7.139440411 x 10

x = 7.139511806 × 10

7.139511806 × 10

1.29 × 10

Therefore:

50.0 g is urea

50.0 g is cinnamic

urea: 50.0 g / 60.06 g/mol = 0.8325 mol

cinnamic acid: 50.0 g / 148.16 g/mol = 0.3375 mol

0.3375 mol / 1.1700 mol = 0.2885

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