Question

The weight in g, of the non-volatile solute urea$\left(N{H}_{2}CON{H}_2\right)$ to be dissolved in $100$ g of water in order to decreases its vapour pressure by $1.8\%$ is.

• A. $0.3$
• B. $3.0$
• C. $0.6$
• D. $6.0$

Answer 1

Answer: (D)

$6.0$

The molar mass of urea is 60 g/mol
Let w g of urea is dissolved in 100 g of water.
The number of moles of urea is $n=\frac{W}{60}$

The number of moles of water is $\frac{100}{18}=5.56$

The mole fraction of urea is
$\frac{\frac{W}{60}}{\frac{W}{60}+5.56}$

The relative lowering in the vapour pressure is equal to the mole fraction of the solute.

$\frac{1.8}{100}=\frac{\frac{W}{60}}{\frac{W}{60}+5.56}$

$0.9259W=\frac{W}{60}+5.56$

$5.56=0.909W$

$W=6.0g$