The mathematical expression for depression in freezing point is given as:
[tex]\Delta T_{f} =i K_{f}\times m[/tex]
where,
[tex]\Delta T_{f}[/tex] = depression in freezing point
[tex]K_{f}[/tex] = molal depression constant ([tex]1.86^{o}C/m[/tex])
m = molality
i = Van't Hoff factor
[tex]\Delta T_{f} =T_{solvent}-T_{solution}[/tex]
= [tex]0.0^{o}C-(-11.2^{o}C)[/tex]
= [tex]11.2^{o}C[/tex]
Now, put the values in formula,
[tex]11.2^{o}C =2\times 1.86^{o}C/m\times m[/tex] (as sodium chloride dissociate into two ions, i =2)
m = [tex]\frac{11.2^{o}C}{1.86^{o}C/m}[/tex]
= [tex]6.02 m[/tex]
Now, molality of the solution = [tex]\frac{number of moles of solute}{kg of the solvent}[/tex]
Number of moles =[tex]\frac{given mass in g}{molar mass}[/tex]
Molar mass of sodium chloride = 58.44 g/mol
molality of the solution = [tex]\frac{\frac{m in g}{58.44 g/mol}}{1.42 kg}[/tex] (1 L = 1kg)
6.02 m = [tex]\frac{\frac{m in g}{58.44 g/mol}}{1.42 kg}[/tex]
mass in g = [tex]6.02 mol/kg \times 58.44 g/mol \times 1.42 kg [/tex]
= [tex]499.56 g [/tex]
Thus, mass of sodium chloride is [tex]499.56 g [/tex]
