Question

On dissolving $10.8g$ glucose (molar weight = $180g/mol$) in $240g$ of water, its boiling point increases by ${0.13}^{\circ }C$. Calculate the molal elevation constant of water.

• A. ${0.75}^{\circ }Ckgmo{l}^{-1}$
• B. ${0.52}^{\circ }Ckgmo{l}^{-1}$
• C. ${0.45}^{\circ }Ckgmo{l}^{-1}$
• D. $1^{\circ }Ckgmo{l}^{-1}$

Answer 1

The correct option is B ${0.52}^{\circ }Ckgmo{l}^{-1}$

Boiling point elevation of a solution is given by,
$△T_b=K_b\times m$
where,
$△T_b$ is the elevation in boiling point.
$K_b$ is molal elevation constant.
m is molality of solution.

$\text{Molality (m)}=\frac{n_{\text{solute}}\times 1000}{W_{\text{solvent}}\text{in gram}}$
$\text{Molality (m)}=\frac{w_{\text{solute}}\times 1000}{M_{\text{solute}}.W_{\text{solvent}}\text{in gram}}$
where,
$w_{\text{solute}}$ is mass of solute
$W_{\text{solvent}}$ is mass of solvent
$M_{\text{solute}}$ is molar mass of solute
$\therefore$
$△T_b=\frac{1000K_b\times w}{W\times M}\text{or}{K}_b=\frac{△T_b\times W\times M}{1000\times w}\text{Given,}△T_b={0.13}^{\circ }C,W=240g,M=180\text{and}w=10.8g{K}_b=\frac{0.13\times 240\times 180}{1000\times 10.8}={0.52}^{\circ }Ckgmo{l}^{-1}$