Question

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

Answer 1

$\pi =CRT$

$2.47=C\times 0.0821\times 303$

$\therefore C=9.93\times {10}^{-2}$

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

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

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

$\therefore$ Mass of water = $\frac{1000}{18.10}\times 18$ = 994.475 gm

Thus, molality of solution $=\frac{9.93\times {10}^{-2}}{994.475\times {10}^{-3}}=9.985\times {10}^{-2}M$

As we know ,

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

putting values , we get

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

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