Question

Addition of 0.643 gram of a compound to 50 ml of benzene $\left( {density\,\,0.879\;g/mL} \right)$ lowers the freezing point from ${5.51^0}C\,\,to\,\,{5.03^0}C$. If ${K_f}$ for benzene is 5.12 calculate the molecular mass of the compound.
(A) 126
(B) 166
(C) 156
(D) 146

Answer 1

Hint: Freezing – point depression is the decrease of the freezing point of a solvent on the addition of non – volatile solute.
To calculate the depression freezing point$ = \Delta {T_f} = {T_f} - T_f^0$
$\Delta {T_f} = $ depression freezing point
${T_f} = $ freezing point of the solution
$T_f^0 = $ freezing point of the pure salt
Density is defined as mass per unit volume
$d = \dfrac{M}{V}$
d = density
M = mass
V = volume
Use this formula to calculate molecular mass of solute
${M_B} = \dfrac{{Kfx{W_B} \times 1000}}{{\Delta Tf \times {W_A}}}$

Complete answer:
Weight of solute $\left( {{W_B}} \right) = 0.643\,\,K.$
Molecular weight of solute $\left( {{M_B}} \right) = ?$
Density $\left( d \right) = 0.879\,\,g/ml$
Freezing point of solution$\left( {{T_f}} \right) = {5.03^0}C$
Freezing point of pure solvent $\left( {{T^0}f} \right) = {5.51^0}C$
Depression constant $\left( {{K_f}} \right) = 5.12$
Volume$\left( V \right) = 50\,\,ml$
Weight of benzene$ = ?$
Weight of solvent $\left( {W_A} \right)$
$d = \dfrac{M}{V}$
$0.879 = \dfrac{M}{{50}}$
Mass of benzene$ = 50 \times 0.879$
$ = 43.95\,\,gram$
$\Delta {T_f} = {T^0}_f - {T_f}$
$\Delta {T_f} = 5.51 - 5.03$
$\Delta {T_f} = {0.48^0}C$
$\Delta {T_f} = {K_f} \times m$
$\Delta {T_f} = Depression\,\,in\,\,F.P$
$m = molality$
${K_f} = Depression\,\,cons\tan t$
$m = \dfrac{{{W_B} \times 1000}}{{{M_B} \times {W_A}\,\,in\,\,gram}}$
$\Delta {T_f} = \dfrac{{Kf \times {W_B} \times 1000}}{{{M_B} \times {W_A}}}$
$0.48 = \dfrac{{5.12 \times 0.643 \times 1000}}{{{M_B} \times 43.9}}$
${M_B} \times 43.9 \times 0.48 = 5.12 \times 0.643 \times 1000$
${M_B} = \dfrac{{5.12 \times 0.643 \times 1000}}{{43.9 \times 0.48}}$
${M_B} = \,156.06\,g/mol.$

Hence correct option is (C).

Note: Freezing point of pure solvent is always greater than freezing point of solution. Depression in freezing point is a colligative property. So, the depression freezing point is increased when we use electrolyte as a solute.