Question

When of alanine are dissolved in of a certain mystery liquid , the freezing point of the solution is less than the freezing point of pure . Calculate the mass of potassium bromide that must be dissolved in the same mass of to produce the same depression in freezing point. The van't Hoff factor for potassium bromide in .

Answer 1

The question is incomplete, the complete question is:

When 177. g of alanine $(C_3H_7NO_2)$ are dissolved in 800.0 g of a certain mystery liquid X, the freezing point of the solution is $5.9^oC$ lower than the freezing point of pure X. On the other hand, when 177.0 g of potassium bromide are dissolved in the same mass of X, the freezing point of the solution is $7.2^oC$ lower than the freezing point of pure X. Calculate the van't Hoff factor for potassium bromide in X.

Answer: The van't Hoff factor for potassium bromide in X is 1.63

Explanation:

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

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

OR

$\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}$ ......(1)

• When alanine is dissolved in mystery liquid X:

$\Delta T_f=5.9^oC$

i = Vant Hoff factor = 1 (for non-electrolytes)

$K_f$ = freezing point depression constant

$m_{solute}$ = Given mass of solute (alanine) = 177. g

$M_{solute}$ = Molar mass of solute (alanine) = 89 g/mol

$w_{solvent}$ = Mass of solvent = 800.0 g

Putting values in equation 1, we get:

$5.9=1\times K_f\times \frac{177\times 1000}{89\times 800}\\\\K_f=\frac{5.9\times 89\times 800}{1\times 177\times 1000}\\\\K_f=2.37^oC/m$

• When KBr is dissolved in mystery liquid X:

$\Delta T_f=7.2^oC$

i = Vant Hoff factor = ?

$K_f$ = freezing point depression constant = $2.37^oC/m$

$m_{solute}$ = Given mass of solute (KBr) = 177. g

$M_{solute}$ = Molar mass of solute (KBr) = 119 g/mol

$w_{solvent}$ = Mass of solvent = 800.0 g

Putting values in equation 1, we get:

$7.2=i\times 2.37\times \frac{177\times 1000}{119\times 800}\\\\i=\frac{7.2\times 119\times 800}{2.37\times 177\times 1000}\\\\i=1.63$

Hence, the van't Hoff factor for potassium bromide in X is 1.63