Question

Calculate the freezing point of a solution containing 8.1g of HBr in 100g of water, assuming the acid to be 90% ionized.
[Given : Molar Mass Br = 80g/mol , Kf water = 1.86 K kg/mol ]

Answer 1

Answer:The freezing point of the solution is $-4^o C$.

Explanation

Weight of solvent 100 g = 0.1 kg (1 kg=1000 g)

Molar mass of HBr = 81 g/mol

Mass of HBr added = 8.1 g

$\Delta T_f=i\times K_f\times \frac{\text{mass of HBr}}{\text{molar mass of HBr}\times \text{weight of solvent in kg}}$

1 = Van'T Hoff factor =

$\alpha =\frac{i-1}{n-1}=90\%=0.9=\frac{i-1}{2-1}$

i = 1.9

$\Delta T_f=1.9\times 1.86 K kg/mol\frac{8.1 g}{81 g/mol\times 0.1kg}=3.534 K$

$\Delta T_f=3.534 K=T^{o}_f-T_f=273 K-T_f$

$T_f=273 K-3.534 K=269.466 K=-4^o C,(T^oC=T(K)-273 K)$

The freezing point of the solution is $-4^o C$.