Question

calculate the freezing point of 3.46 gram of a compound X in 160 gram of benzene when a separate sample of X was vaporized it's density was found to be 3.27 gram/liter at 116°c and 773 torr. The freezing point of pure benzene is 5.45°c of Kf=5.1°/m​

Answer 1

Answer: The freezing point of 3.46 gram of a compound X in 160 gram of benzene is $4.38^0C$

Explanation:

The relation of density and molar mass is:

$d=\frac{PM}{RT}$

where

d = density = 3.27 g/ L

P = pressure of the gas  = 773 torr = 1.02 atm   (760 torr = 1atm)

M = molar mass of the gas  = ?

T = temperature of the gas = $116^0C=(116+273)K=389K$

R = gas constant  = $0.0821Latm/Kmol$

$M=\frac{dRT}{P}=\frac{3.27g/L\times 0.0821Latm/Kmol\times 389K}{1.02atm}=102.3g/mol$

The relation of depression in freezing point with molality:

$\Delta T_f=k_f\times m$

$\Delta T_f$ = depression in freezing point = $T_f^0-T_f$ = $5.45-T_f$

$k_f$ = freezing point constant  = 5.1

m = molality = $\frac{\text {moles of X}}{\text {weight of solvent in kg}}=\frac{3.46\times 1000}{102.3\times 160}=0.21$

$5.45-T_f=5.1\times 0.21$

$T_f=4.38^0C$

Thus the freezing point of 3.46 gram of a compound X in 160 gram of benzene is $4.38^0C$