Question

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.

Answer 1

Answer:

$\large\boxed{\large\boxed{107g}}$

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°.

               $\Delta P=X_{solute}\times P^0$

ΔP and P⁰ are given:

• ΔP = 1.39mmHg

• P⁰ = 55.32mmHg

1. Mole fraction of solute

Thus, you can calculate Xsolute:

           $1.39mmHg=X_{solute}\times 55.32mmHg\\\\X_{solute}=0.02513$

2. Moles of solute

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

• Number of moles of water = mass in grams / molar mass

• Number of moles of water = 1,250g/18.015 (g/mol) = 69.3866mol

• Xsolvent = 1 - Xsolute = 1 - 0.02513 = 0.97487

• Xsolvent = moles of solvent / moles of solution

• 0.97487 = 69.3866mol / moles of solution

• moles of solution = 71.1752mol

• moles of solute = moles of solution - moles of solvent = 71.1752mol - 69.3866mol = 1.7886mol

3. Mass of urea

Formula:

• Mass = number of moles × molar mass

• Mass = 1.7886 mol × 60.06 g/mol = 107.42 g

You must round to 3 significant figures: 107 g