Question

This problem has been solved!See the answerWhen 282. g of glycine (C2H5NO2) are dissolved in 950. g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 282. g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for sodium chloride in X.

Answer 1

Answer:

The van't Hoff factor for sodium chloride in X is 1.9 .

Explanation:

Mass of liquid x = 950 g = 0.950 kg

When 282 g of glycine are dissolved in 950 g of a certain mystery liquid X.

Depression in freezing point of the solution $\Delta T_f= 8.2^oC$

Molality of the solution = m

The van't Hoff factor for glycine (non ionic) in liquid X= i =1

$m=\frac{282 g}{75.07 g/mol\times 0.950 kg}=3.9542 mol/kg$

$\Delta T_f=i\times K_f\times m$

$8.2^oC=1\times K_f\times 3.9542 mol/kg$

The value of molal depression constant for liquid X:

$K_f=2.0737 ^oC kg/mol$..(1)

Now 282. g of sodium chloride are dissolved in the same mass of X.

Depression in freezing point of the NaCl solution $\Delta T_f'= 20.0^oC$

Molality of the NaCl solution = m'

The van't Hoff factor for NaCl(ionic) in liquid X= i'

$m'=\frac{282 g}{58.5 g/mol\times 0.950 kg}=5.0742 mol/kg$

$\Delta T_f'=i'\times K_f\times m'$

$20.0^oC=i'\times 2.0737 ^oC kg/mol\times 5.0742 mol/kg$

i = 1.9007 ≈ 1.9

The van't Hoff factor for sodium chloride in X is 1.9 .