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Question

18 g of glucose (C6H12O6) is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at 100C is

A
759.00 torr
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B
7.60 torr
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C
76.00 torr
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D
752.40 torr.
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Solution

The correct option is D 752.40 torr.
Relative lowering of vapour pressure is equal to mole fraction of glucose
PoPPo= mole fraction of glucose...(i) moles of glucose=18180=0.1 moles of water=178.218=9.9mole fraction of glucose=0.10.1+9.9=0.01
Substituting in equation (i) we get,
760P760=0.01
On solving, we get,
P=752.4 torr

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