The molal lowering of vapour pressure for water at 100 o C , is, 760 mm, 750 mm, 13.43 mm, 0.760 mm

P 0 - P S P 0 = n 2 n 1 + n 2 P 0 - P S = 1 1 + 1000 18 × 760 = 13.43 m m

The molal lowering of vapour pressure for water at 100 o C , is

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The molal lowering of vapour pressure for water at

100^{o} C

, is

ChemistrySolutionsNEET

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Solution:

2696 Upvotes Verified Answer

The correct answer is: 13.43 mm

\frac{P^{0} - P_{S}}{P^{0}} = \frac{n_{2}}{n_{1} + n_{2}}

P^{0} - P_{S} = [\frac{1}{1 + \frac{1000}{18}}] \times 760 = 13.43 m m

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