Question

When of glycine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezing point of pure . On the other hand, when of ammonium chloride are dissolved in the same mass of , the freezing point of the solution is lower than the freezing point of pure . Calculate the van't Hoff factor for ammonium chloride in .

Answer 1

The given question is incomplete. The complete question is:

When 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 ammonium 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 ammonium chloride in X.

Answer:  the van't Hoff factor for ammonium chloride is 1.74

Explanation:

Depression in freezing point is given by:

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

$\Delta T_f=T_f^0-T_f=8.2^0C$ = Depression in freezing point  

$K_f$ = freezing point constant = ?

i = 1 ( for non electrolyte)

m= molality

$8.2^0C=1\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}$

Weight of solvent (X)= 950 g = 0.95 kg

Molar mass of solute (glycine) = 75.07 g/mol

Mass of solute (glycine) = 282 g

$8.2^0C=1\times K_f\times \frac{282g}{75.07g/mol\times 0.95kg}$

$K_f=2.07$

ii) $20.0^0C=i\times \times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}$

Weight of solvent (X)= 950 g = 0.95 kg

Molar mass of solute (ammonium chloride) = 53.49 g/mol

Mass of solute (ammonium chloride) = 282 g

$20.0^0C=i\times 2.07\times \frac{282g}{53.49g/mol\times 0.95kg}$

$i=1.74$

Thus the van't Hoff factor for ammonium chloride is 1.74