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Question: Addition of 0.643 gram of a compound to 50 ml of benzene \(\left( {density\,\,0.879\;g/mL} \right)\)...

Addition of 0.643 gram of a compound to 50 ml of benzene (density  0.879  g/mL)\left( {density\,\,0.879\;g/mL} \right) lowers the freezing point from 5.510C  to  5.030C{5.51^0}C\,\,to\,\,{5.03^0}C. If Kf{K_f} for benzene is 5.12 calculate the molecular mass of the compound.
(A) 126
(B) 166
(C) 156
(D) 146

Explanation

Solution

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=ΔTf=Tf−Tf0 = \Delta {T_f} = {T_f} - T_f^0
ΔTf=\Delta {T_f} = depression freezing point
Tf={T_f} = freezing point of the solution
Tf0=T_f^0 = freezing point of the pure salt
Density is defined as mass per unit volume
d=MVd = \dfrac{M}{V}
d = density
M = mass
V = volume
Use this formula to calculate molecular mass of solute
MB=KfxWB×1000ΔTf×WA{M_B} = \dfrac{{Kfx{W_B} \times 1000}}{{\Delta Tf \times {W_A}}}

Complete answer:
Weight of solute (WB)=0.643  K.\left( {{W_B}} \right) = 0.643\,\,K.
Molecular weight of solute (MB)=?\left( {{M_B}} \right) = ?
Density (d)=0.879  g/ml\left( d \right) = 0.879\,\,g/ml
Freezing point of solution(Tf)=5.030C\left( {{T_f}} \right) = {5.03^0}C
Freezing point of pure solvent (T0f)=5.510C\left( {{T^0}f} \right) = {5.51^0}C
Depression constant (Kf)=5.12\left( {{K_f}} \right) = 5.12
Volume(V)=50  ml\left( V \right) = 50\,\,ml
Weight of benzene=? = ?
Weight of solvent (WA)\left( {W_A} \right)
d=MVd = \dfrac{M}{V}
0.879=M500.879 = \dfrac{M}{{50}}
Mass of benzene=50×0.879 = 50 \times 0.879
=43.95  gram= 43.95\,\,gram
ΔTf=T0f−Tf\Delta {T_f} = {T^0}_f - {T_f}
ΔTf=5.51−5.03\Delta {T_f} = 5.51 - 5.03
ΔTf=0.480C\Delta {T_f} = {0.48^0}C
ΔTf=Kf×m\Delta {T_f} = {K_f} \times m
ΔTf=Depression  in  F.P\Delta {T_f} = Depression\,\,in\,\,F.P
m=molalitym = molality
Kf=Depression  constan⁡t{K_f} = Depression\,\,cons\tan t
m=WB×1000MB×WA  in  gramm = \dfrac{{{W_B} \times 1000}}{{{M_B} \times {W_A}\,\,in\,\,gram}}
ΔTf=Kf×WB×1000MB×WA\Delta {T_f} = \dfrac{{Kf \times {W_B} \times 1000}}{{{M_B} \times {W_A}}}
0.48=5.12×0.643×1000MB×43.90.48 = \dfrac{{5.12 \times 0.643 \times 1000}}{{{M_B} \times 43.9}}
MB×43.9×0.48=5.12×0.643×1000{M_B} \times 43.9 \times 0.48 = 5.12 \times 0.643 \times 1000
MB=5.12×0.643×100043.9×0.48{M_B} = \dfrac{{5.12 \times 0.643 \times 1000}}{{43.9 \times 0.48}}
MB= 156.06 g/mol.{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.