SolutionHard
Question
Lowering of vapour pressure in 1 molal aqueous solution at 100°C is
Options
A.13.44 mm Hg
B.14.12 mm Hg
C.31.2 mm Hg
D.35.2 mm Hg
Solution
$P^{o} - P = X_{1}.P^{o} = \frac{1}{1 + \frac{1000}{18}} \times 760 = 13.44\text{ mm kg}$
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