SolutionHard
Question
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 :
Options
A.3 mol
B.2 mol
C.1 mol
D.0.5 mol
Solution
Lowering of V.P. is colligative property
thus, iK2SO4 = 1 + (y - 1)x = 1 + 2x = 3
∴ If
n1 = 1
thus, iK2SO4 = 1 + (y - 1)x = 1 + 2x = 3
∴ If
n1 = 1
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