Answer to Question #150144 in General Chemistry for heni

Question

A solution contains 10.0 g of carbon tetrachloride, CCl4, and 20.0 g of benzene, C6H6. Calculate the mole fraction of each component.

Answer 1

MM(CCl₄) = 153.82 g/mol

n(CCl₄) $= \frac{10.0}{153.82} = 0.065 \;mol \\$

MM(C₆H₆) = 78.11 g/mol

n(C₆H₆) $= \frac{20.0}{78.11} = 0.256 \;mol$

Total moles = 0.065 + 0.256 = 0.321 mol

Proportion:

0.321 – 1

0.065 – x

$x = \frac{0.065 \times 1}{0.321} = 0.2025$

The mole fraction of CCl₄ is 0.2025

The mole fraction of C₆H₆ is (1 – 0.2025) = 0.7975