Question

Calculate the molal depression constant of a solvent which has freezing point ${\text{16}}{\text{.}}{{\text{6}}^{\text{o}}}{\text{C}}$and latent heat of fusion${\text{180}}{\text{.75J/g}}$.
(A) 2.68
(B) 3.86
(C) 4.68
(D) 2.86

Answer 1

Hint: The molal depression constant (${K_f}$) can be found from the depression in freezing point, ${K_f}$ and molality ($m$). But it is related to the latent heat of fusion (${l_f}$) inversely. Latent heat of fusion is the change in energy observed in a substance in its solid form converting into liquid form.

Complete step by step solution:
The given values are of freezing point and latent heat of fusion. The thermodynamic relation between latent heat of fusion and molal depression constant is given by the following expression ${K_f} = \dfrac{{R{T_f}^2}}{{{l_f} \times 1000}}$.
${K_f}$- Molal depression constant
$R$- Universal gas constant
${T_f}$- Freezing point of the solvent
${l_f}$-latent heat of fusion
The given value of ${T_f}$ is ${16.6^ \circ }$C that equals to (16.6+273.15= 289.75K), the value of ${l_f}$ is $180.75J/g$, the value of $R$ is $8.314J/mol K$. On substituting the above values in the above equation, $\begin{gathered}
{K_f} = \dfrac{{R{T_f}^2}}{{{l_f} \times 1000}} \\
{K_f} = \dfrac{{8.314 \times {{\left( {289.75} \right)}^2}}}{{180.75 \times 1000}} \\
\Rightarrow {K_f} = 3.8577 \simeq 3.86 \\
\end{gathered} $K kg /mol

The answer to the given question is option (B) 3.86 K kg/mol.

Additional information:
Depression in freezing point ($\Delta {T_f}$) is a colligative property and it is directly proportional to molality. The proportionality constant introduced here is the molal freezing point depression constant$\left( {{K_f}} \right)$. The formula for freezing point depression is $\Delta {T_f} = {K_f}{\kern 1pt} m$. When the molal depression constant is asked in relation to the latent heat of fusion, the above formula can be used. When the molar mass of a solvent is mentioned in the question, then the formula can be taken as ${K_f} = \dfrac{{{M_{{W_A}}}R{T_f}^2}}{{{\Delta _{fus}}H \times 1000}}$.

Note: During the calculation, the value of gas constant (R) is taken as 8.314 J/mol K in the S.I units because the heat of fusion is in Joules. The value of temperature is also converted to the Kelvin scale. To convert the Celsius scale to Kelvin scale, the given value of temperature has to be added with 273.15 kelvin i.e. t$^ \circ $C + 273.15K.