Question

Vapour pressure of water is $360$mm Hg, how much urea should be added to $200$mL water to reduce its vapour pressure by $0.5\%$? (Molecular wt. of urea $=60$)

• A. $2.52g$
• B. $3.33g$
• C. $3.96g$
• D. None of these

Answer 1

Answer: (B)

$3.33g$

Vapour pressure of pure water = $p^o$ = 360 mm Hg

Lowered vapour pressure = p = 360 - 0.5 % of 360 = 360 - 1.8 = 358.2 mm Hg

Now,

weight of water = w1 = $200ml\times 1g/ml$ = 200 g

Weight of urea = w2 = ?

Molecular weight of water M1 = 18 g/mol

Molecular weight of urea M2 = 60 g/mol

According to Raoult's law:

$\frac{p^o-p}{p}=\frac{n_2}{n_1+n_2}$

$\frac{360-358.2}{358.2}=\frac{\frac{w_2}{M_2}}{\frac{w_1}{M_1}+\frac{w_2}{M_2}}$

$0.005=\frac{\frac{w_2}{M_2}}{\frac{w_1}{M_1}+\frac{w_2}{M_2}}$

$0.005[\frac{w_1}{M_1}+\frac{w_2}{M_2}]=\frac{w_2}{M_2}$

$0.005\times \frac{w_1}{M_1}=\frac{w_2}{M_2}-0.005\times \frac{w_2}{M_2}$

$0.005\times \frac{w_1}{M_1}=0.995\times \frac{w_2}{M_2}$

$\frac{0.005}{0.995}\times \frac{w_1}{M_1}\times M_2=w_2$

$w_2=0.005\times \frac{200}{18}\times 60$

$w_2=3.33g$

Hence, 3.33 g of urea must be added.