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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a
3 mol
b
2 mol
c
1 mol
d
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