Question

The osmotic pressure of an aqueous solution of sucrose is 2.47 atm at 303 K and the molar volume of the water present is $18.10c{m}^3$. Given $\Delta H_{vap}=540cal/g$. Assume the volume of solvent equal to the volume of the solution. The elevation in the boiling point of the solution is :

• A. $5.145\times {10}^{-2}$
• B. $7.565\times {10}^{-2}$
• C. $6.355\times {10}^{-2}$
• D. None of these

Answer 1

Answer: (A)

$5.145\times {10}^{-2}$

$\pi =CST$
$2.47=C\times 0.0821\times 303$
$\therefore C=9.93\times {10}^{-2}M$

Thus, 1 litre solution of sucrose contains $9.93\times {10}^{-2}$ mole of sucrose or $9.93\times {10}^{-2}\times 342$ g of sucrose.

$\because$ Volume of solution $=$ Volume of solvent $=1000mL$

$\therefore$ Mole of water $=\frac{1000}{18.10}$

$\therefore$ Mass of water $=\frac{1000}{18.10}\times 18=994.475g$
Thus, molality of solution $=\frac{9.93\times {10}^{-2}}{994.475\times {10}^{-3}}=9.985\times {10}^{-2}M$

$\Delta T_b=K_b\times molality=\frac{R{T}_b^2}{1000l}\times molality$

$\Delta T_b=\frac{2\times 373\times 373}{1000\times 540}\times 9.985\times {10}^{-2}$

$\Delta T_b=5.145\times {10}^{-2}$