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 $4.38^0C$

Answer 2

From the information contained in the question, the freezing point of the solution is 4.38°C.

The molar mass of the compound can be obtained from;

M = dRT/P

Where;

d = Density of the gas = 3.27 gram/liter

R = molar constant =  0.082 atmL-K-1mol-1

T = Absolute temperature = 389 K

P = pressure = 773 torr or 1.02 atm

M = 3.27 gram/liter × 0.082 atmL-K-1mol-1 × 389 K/1.02 atm

M = 102.3 g/mol

Number of moles of X =  3.46 gram/102.3 g/mol = 0.034 moles

Molality of the solution = Number of moles of solute/ Mass of solution in kilograms = 0.034 moles/0.16 = 0.21 m

Using the relation;

ΔT = K m i

Where;

ΔT = Freezing point depression

K = Freezing constant = 5.1°C/m​

m = molality of solution = 0.21 m

i = Van't Hoff factor = 1

ΔT =  5.1°C/m​ × 0.21 m × 1

ΔT = 1.07 °C

But;

ΔT = Freezing point of pure solvent - Freezing point of solution

Freezing point of solution =  Freezing point of pure solvent - ΔT

Freezing point of pure solvent = 5.45°C

Freezing point of solution =   5.45°C - 1.07 °C

Freezing point of solution = 4.38°C

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