Question

An aqeous solution of urea is made by dissolving 10 g of urea in 90 g water at 303k. if the v.p of pure water at 303k be 32. 8mm kg . what would be v.p of the solution?

Answer 1

According to Raoult's law,

$P_{solution} =$χ$_{solvent} P_{solvent} ^{0}$

Calculating the molefraction of water (solvent):

Moles of urea = $10 g (NH_{2})_{2}CO *\frac{1 mol }{60.06 g } = 0.167 mol (NH_{2})_{2}CO$

Moles of water = $90 g H_{2}O * \frac{1 mol H_{2}O}{18 g H_{2}O} = 5 mol H_{2}O$

Molefraction of solvent = $\frac{5 mol}{(5 + 0.167)mol}$ = 0.968

Vapor pressure of the solution = χ$_{solvent} P_{solvent} ^{0}$

= 0.968 (32.8 mmHg)

= 31.7 mmHg

Answer 2

Explanation:

hope it will help you !!