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Question
Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
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Solution
Given: Mass of benzene in the solution = 30 g
Let the mass of the solution = 100 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 benzene (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
Mole fraction of benzene (C6H6) = `"Number of moles of benzene"/("Number of moles of benzene" + "Number of moles of Carbon tetrachloride")`
= `0.385/(0.385 + 0.455)`
= `0.385/0.84`
= 0.458
∴ Mole fraction of CCl4 = 1 − 0.458
= 0.542
∴ The mole fraction of benzene in a solution containing 30% by mass of carbon tetrachloride is 0.542.
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