Question

1 g mixture of glucose and urea present in 250 mL aqueous solution shows the osmotic pressure of 0.74 atm at ${27}^{o}C$. Assuming solution to be diluted, which are correct?

1. Percentage of urea in the mixture is 17.6.
2. Relative lowering of the vapour pressure of this solution is $5.41\times {10}^{-4}$.
3. The solution will boil at ${100.015}^{o}C$, if $K_b$ of water is 0.5 K molality${}^{-1}$.
4. If glucose is replaced by the same amount of sucrose, the solution will show higher osmotic pressure at ${27}^{o}C$.
5. If glucose is replaced by the same amount of $NaCl$, the solution will show lower osmotic pressure at ${27}^{o}C$.

• A. 1, 2, 3
• B. 1, 2, 3, 5
• C. 2, 4, 5
• D. 1, 4, 5

Answer 1

Answer: (A)

1, 2, 3

Let a g glucose, b g urea be present in 1 g

$\therefore a+b=1$ ....(i)

Thus, $0.74\times \frac{250}{1000}=[\frac{a}{180}+\frac{b}{60}]\times 0.0821\times 300$ ....(ii)

By Eqs. (i) and (ii), $b=0.176,a=0.824\therefore \%Urea$ in 1 g $=17.6\%$

$\frac{P^0-P_S}{P^0}=\frac{n_{urea}{+}_{glucose}}{n_{water}}$

$\therefore \frac{P^0-P_S}{P^0}=\frac{0.0029+0.0046}{13.89}=5.4\times {10}^{-4}$

$\Delta T_b=$molality $\times K_b=\frac{X_B\times 1000}{M_{water}}\times K_b$

$=\frac{5.4\times {10}^{-4}\times 1000}{18}\times 0.5=0.015$

$\therefore \text{Boiling point}=100+0.015={100.015}^{o}C$

On replacing glucose (Molar mass 180) by sucrose (Molar mass 342), $\pi$ will decreases as $\pi \propto \frac{1}{\text{Molar mass}}$.

On replacing glucose (Molar mass 180) by NaCl (Molar mass 585), $\pi$ will increase as $\pi \propto \frac{1}{\text{Molar mass}}\times 2$ for NaCl.

Hence, the correct option is $\text{A}$

Option A is correct.