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Question
Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
Solution
Let the total mass of the solution be 100 g, and the mass of benzene be 30 g.
∴ Mass of carbon tetrachloride = (100 − 30) g
= 70 g
Molar mass of benzene (C6H6) = (6 × 12 + 6 × 1) g mol−1
= 78 g mol−1
∴ Number of moles of C6H6 = `"Mass"/"Molar mass"`
= `(30 "g")/(78 "g mol"^(−1))`
= 0.385 mol
Molar mass of carbon tetrachloride (CCl4) = 1 × 12 + 4 × 35.5
= 154 g mol−1
∴ Number of moles of CCl4 = `"Mass"/"Molar mass"`
= `(70 "g")/(154 "g mol"^(−1))`
= 0.455 mol
Thus, the mole fraction of C6H6 is given as:
`("Number of moles of C"_6"H"_6)/("Number of moles of C"_6"H"_6+"Number of moles of CCl"_4)`
= `0.385/(0.385+0.455)`
= 0.458
∴ Mole fraction of CCl4 = 1 − 0.458 = 0.542
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