Question

The freezing point of p-dichlorobenzene (MW = 146.992 g/mol) is 53.1 °C. Its Kf value is 7.10 °C/m. A solution of 1.52 g of sulfanilamide in 10.0 g of p-dichlorobenzene freezes at 46.7 °C. ?

If the vapor pressure of pure p-dichlorobenzene is 10.0 mmHg at 54.8 °C, what is the vapor pressure of the sulfanilamide solution at 54.8 °C?
The answer is 8.83 mmHg but I don't know how to get to it....

Answer 1

`"8.83 mmHg"`

Explanation:

On adding `"1.52 g"` of Sulfanilamide (solute) to `"10.0 g"` of p-dichlorobenzene (solvent) the freezing point of solvent decreases from 53.1°C to 46.7°C
This depression in freezing point is given as

`DeltaT_f = -iK_fm`

Where

• `i =` Van't Hoff factor of solvent

• `K_f =` Freezing point constant of solvent

• `m = "molality of solution" = "moles of solute"/"mass of solvent (in kg)"`

`("46.7°C - 53.1°C") = -i × "7.10°C/m" × ("1.52 g")/("172.2 g/mol" × 10.0 xx 10^-3 "kg")`

Solving above equation will give us `i = 1.02`

Now,
Relative lowering of vapour pressure is given by,

`(P° - P)/(P°) = iX`

Where

• `P° =` Vapour Pressure of pure solvent

• `P =` Vapour Pressure of solution

• `i =` Van't Hoff factor of solvent

• `X =` Mole fraction of solute

`("10.0 mmHg " - P)/("10.0 mmHg") = 1.02 × ("1.52g"/"172.2 g/mol")/("1.52g"/"172.2 g/mol" + "10.0 g"/"146.992 g/mol")`

Solving above equation will give us `P = "8.83 mmHg"`