How many grams of urea (non volatile, non electrolyte) must be added to 1250. g water to give a solution with a vapor pressure lowering of 1.39 mm Hg at 40.0 degree C? Water pressure is 55.32 mm Hg at this temp. Urea is (NH 2) 2CO.

Explanation:

The vapor pressure lowering is a colligative property and it follows  Raoult's law.

The vapor pressure lowering of a solvent in a solution, ΔP, is equal to the mole fraction of the solute, Xsolute, multiplied by the vapor pressure of the pure solvent, P°.

ΔP and P⁰ are given:

1. Mole fraction of solute

Thus, you can calculate Xsolute:

2. Moles of solute

Now you can calculate the number of moles of water and the number of moles of solute.

3. Mass of urea

Formula:

You must round to 3 significant figures: 107 g

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