Question

The freezing point of 1% aqueous solution of calcium nitrate will be?

Answer 1

less than 0 is the answer

Answer 2

Answer: 272.89 K

Explanation: We are given that $Ca(NO_3)_2$ is present in 1 % solution, which means that 1.0 gram of calcium nitrate is present in 100 gram of the solution.

$\text{Number of moles of solute}=\frac{\text{Given mass}}{\textMolar mass}}$

$\text{Number of moles of solute}=\frac{1g}{164g/mol}=0.006moles$

Mass of solvent = mass of solution - mass of solute = 100 -1 = 99 g= 0.099 kg

Depression in freezing point :

Formula used for lowering in freezing point is,

$\Delta T_f=k_f\times m$

or,

$\Delta T_f=k_f\times \frac{\text{ Moles of solute}}{\text{ Mass of solvent in kg}}$

where,

$T_f$ = change in freezing point

$k_f$ = freezing point constant

m = molality

$\Delta T_f=1.86 \times \frac{0.006}{0.099 kg}=0.11$

$T^{o}_f-T_f=273 K-T_f=0.11$

$T_f=272.89K$