Question

Calculate the mass of urea required to be dissolved in $100g$ of water, so as the vapour pressure of the solution decreases by $50\%$.
Given: $\text{Molecular mass of urea is}60gmo{l}^{-1}$

• A. $33.33g$
• B. $22.22g$
• C. $222.22g$
• D. $333.33g$

Answer 1

The correct option is D $333.33g$

Using Raoult's law in the following form for non dilute solution:
$\frac{P_A^0-P_s}{P_s}=\frac{w_B\times M_A}{w_A\times M_B}$
$w_A=\text{mass of A (Solvent)}$
$w_B=\text{mass of B (Solute)}$
$M_A=\text{molar mass of A}$
$M_B=\text{molar mass of B}$

If $P_A^0=100mm$, then $P_s=50mm$
Putting the values:

$\frac{100-50}{50}=\frac{w_B\times 18}{100\times 60}$

$w_B=333.33g$