Moles of K2SO4 to be dissolved in 12 mol water to lower its vapour pressure by 10 mmHg at a temperature at which vapour pressure of pure water is 50 mm is:

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3 mol

2 mol

1 mol

0.5 mol

answer is C.

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Detailed Solution

Lowering of V.P. is colligative property thus, iK2SO4=1+(n−1)α=1+3-11=3∴ If Δppo=n1in1i+n21050=3n13n1+12 15=n1n1+4 n1=1

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