Question

When 74.8g of alanine C3H7NO2 are dissolved in 1450.g of a certain mystery liquid X, the freezing point of the solution is 8.30°C less than the freezing point of pure X.Calculate the mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point. The van't Hoff factor =i1.72 for potassium bromide in X.

Answer 1

Answer: The mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

Explanation:

Depression in freezing point is given by:

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

$\Delta T_f=T_f^0-T_f=8.30^0C$ = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte)

$K_f$ = freezing point constant =  

m= molality = $\frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}=\frac{74.8g\times 1000}{1450g\times 89.09g/mol}=0.579$

$8.30^0C=1\times K_f\times 0.579$

$K_f=14.3^0C/m$

Let Mass of solute (KBr) = x g

$8.3^0C=1.72\times 14.3\times \frac{xg\times 1000}{119g/mol\times 1450g}$

$x=58.2g$

Thus the mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams