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Question

# When a liquid that is immiscible with water was steam distilled at 95.2oC at a total pressure of 99.652 kPa, the distillate contained 1.27 g of the liquid per gram of water. What will be the molar mass of the liquid if the vapour pressure of water is 85.140 kPa at 95.2oC?

A
99.65 gmol1
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B
18 gmol1
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C
134.1 gmol1
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D
105.74 gmol1
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Solution

## The correct option is C 134.1 gmol−1Total pressure, Ptotal=99.652 kPaPwater=pB=85.140 kPaPliquid=pA=(99.652−85.140) kPa=14.512 kPaand mAmB=1.271or mAmB=pAMApBMBor, MA=(mAmB)(pBMBpA)MA=1.27×(85.140kPa×18 g mol−114.512 kPa)MA≃134.1 g mol−1

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