Question

Calculate the vapour pressure for 45 w/w aqueous solution of urea of 298 K temperature. The vapour pressure
of water is 0.025 bar.​

Answer 1

0.25 molar aqueous solution of urea means: 1000 g of ... (Vapor pressure of pure water at 298 K is 23.8 mm Hg) ... Now, we have to calculate vapour pressure of water in the solution.

Answer 2

The vapor pressure of the solution at 298 K is 0.020 bar.

Explanation:

As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.

The formula for relative lowering of vapor pressure will be,

$\frac{p^o-p_s}{p^o}=i\times x_2$

where,

$\frac{p^o-p_s}{p^o}$= relative lowering in vapor pressure

i = Van'T Hoff factor = 1 (for non electrolytes)

$x_2$ = mole fraction of solute   =$\frac{\text {moles of solute}}{\text {total moles}}$

Given : 45 g of urea is present in 100 g of aqueous solution, thus (100-45) g = 55 g of water

moles of solute (urea) = $\frac{\text{Given mass}}{\text {Molar mass}}=\frac{45g}{60g/mol}=0.75moles$

moles of solvent (water) = $\frac{\text{Given mass}}{\text {Molar mass}}=\frac{55g}{18g/mol}=3.1moles$

Total moles = moles of solute (urea) + moles of solvent (water) = 0.75 + 3.1 = 3.85

$x_2$ = mole fraction of solute  =$\frac{0.75}{3.85}=0.19$

$\frac{0.025-p_s}{0.025}=1\times 0.19$

$p_s=0.020$

Thus the vapor pressure of the solution at 298 K is 0.020 bar.

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