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

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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The freezing point of 3.46 gram of a compound X in 160 gram of benzene is

Explanation:

The relation of density and molar mass is:

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 =

R = gas constant  =

The relation of depression in freezing point with molality:

= depression in freezing point = =

= freezing point constant  = 5.1

m = molality =

Thus the freezing point of 3.46 gram of a compound X in 160 gram of benzene is

the correct answer is:

a. formation of precipitation