Answer to Question #150144 in General Chemistry for heni

Question #150144
A solution contains 10.0 g of carbon tetrachloride, CCl4, and 20.0 g of benzene, C6H6. Calculate the mole fraction of each component.
1
Expert's answer
2020-12-14T14:34:50-0500

MM(CCl4) = 153.82 g/mol

n(CCl4) =10.0153.82=0.065  mol= \frac{10.0}{153.82} = 0.065 \;mol \\

MM(C6H6) = 78.11 g/mol

n(C6H6) =20.078.11=0.256  mol= \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=0.065×10.321=0.2025x = \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


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