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Q.

What weight of glucose must be dissolved in 100g of water to lower the vapour pressure by 0.20 mmHg ?(Assume dilute solution is being formed)
Given : Vapour pressure of pure water is 54.2 mmHg at room temperature. Molar mass of glucose is 180 g mol−1
 

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a

4.69 g

b

3.69 g

c

3.59 g

d

2.59 g

answer is B.

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Detailed Solution

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Po−PsP0=WM.W×M.W(W)H2O 0.254.2=W180×18100 W=3.69

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