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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.

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Answer:

[tex]\large\boxed{\large\boxed{107g}}[/tex]

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°.

               [tex]\Delta P=X_{solute}\times P^0[/tex]

ΔP and P⁰ are given:

  • ΔP = 1.39mmHg
  • P⁰ = 55.32mmHg

1. Mole fraction of solute

Thus, you can calculate Xsolute:

           [tex]1.39mmHg=X_{solute}\times 55.32mmHg\\\\X_{solute}=0.02513[/tex]

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

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