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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 gmol1
Total pressure, Ptotal=99.652 kPa

Pwater=pB=85.140 kPa

Pliquid=pA=(99.65285.140) kPa

=14.512 kPa

and mAmB=1.271

or mAmB=pAMApBMB

or, MA=(mAmB)(pBMBpA)

MA=1.27×(85.140kPa×18 g mol114.512 kPa)

MA134.1 g mol1

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