Question

A soluction contains $7g$ of solute (molar mass $210g$ ${mol}^{-1}$) in $350g$ of acetone raised the boiling point of acetone from ${56}^{o}C$ to $56{.3}^{o}C$.

The value of ebullioscopic constant of acetone in $K.Kg$ ${mol}^{-1}$ is:

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

Answer 1

The correct option is B $3.15$

$T_b=56.3℃=(273+56.3)K=329.3K$

${T_b}^0=56℃=(273+56)K=329K$

Mass of solute $=7g$

Mass of solution $=(350+7)g=357g$

Molality of solution (m) $=\frac{\text{No. of moles of solute}}{\text{kg of solvent}}=\frac{\left(\frac{\text{mass of solute}}{\text{molar mass of solute}}\right)}{\text{Kg of solvent}}$

$\Rightarrow m=\frac{\left(\frac{7}{210}\right)}{\left(\frac{350}{1000}\right)}=\frac{10}{105}$

As we know that, depression in boiling point,

$\Delta T_b=k_b.m$

where,

$\Delta T_b$ = Elevation in boiling point = $T_b-{T_b}^0=(329.3-329)=0.3K$

$k_b$ = ebullioscopic constant of acetone = molal elevation of boiling point = $?$

$m$ = molality of solution = $\frac{10}{105}$

$\therefore 0.3=k_b\times \frac{10}{105}$

$\Rightarrow k_b=0.3\times \frac{105}{10}$

$\Rightarrow k_b=3.15KKg{mol}^{-1}$