Question

A solution containing $7g$ of a solute (molar mass = $210gmo{l}^{-1}$) in $350g$ of acetone raised the boiling point of acetone from ${56}^{\circ }C\text{to}{56.3}^{\circ }C$. The value of Ebullioscopic
constant of acetone in $Kkgmo{l}^{-1}$ is

• A. 2.66
• B. 3.15
• C. 4.12
• D. 2.86

Answer 1

The correct option is B 3.15

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
$\therefore$
$△T_b=\frac{1000K_b\times w}{W\times M}\text{or}{K}_b=\frac{△T_b\times W\times M}{1000\times w}\text{Given,}△T_b=0.3K\left({\text{Change in temperature can be expression in both}}^{\circ }CorK\right)W=350g,M=210\text{and}w=7g$
Substituting the given values, we get,

$0.3=K_b\times \frac{7\times 1000}{210\times 350}{K}_b=\frac{0.3\times 210\times 350}{7\times 1000}=3.15Kkgmo{l}^{-1}$