Question

5 g of Na2SO4 was dissolved in x g of H2O.The change in freezing point was foundto be 3.82C. If Na2SO4 is 81.5% ionised,the value of x

Answer 1

Answer: The value of x is 0.07 kg.

Explanation:-

Weight of solvent (water)= x kg      

Molar mass of  $(Na_2SO_4)$ = 92 g/mol

Mass of $(Na_2SO_4)$ added = 5 g

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

$\Delta T_f$ = change in freezing point= $3.82^0C$

$K_f$ =  freezing point constant =$1.86Kkg/mol$

i = Van'T Hoff factor

$Na_2SO_4\rightarrow 2Na^++SO_4^{2-}$

$\alpha =\frac{i-1}{n-1}=81.5\%=0.9=\frac{i-1}{3-1}$

i = 2.63

$3.82=2.63\times 1.86 K kg/mol\frac{5 g}{92 g/mol\times xkg}$

$x=0.07kg$