Formula for the depression in freezing point is:
[tex]\Delta T_f = i\times k_f\times m[/tex] -(1)
where [tex]\Delta T_f[/tex] is depression in freezing point,
[tex]i[/tex] is Van't Hoff factor,
[tex]k_f[/tex] is molal freezing point depression constant, and
[tex]m[/tex] is molality of the solution.
Molality of the solution, [tex]m = 0.99 m[/tex] (given)
Molal freezing point depression constant of water, [tex]k_f = 1.86^{o}C/m[/tex]
Depression in freezing point of solution, [tex]\Delta T_f = T_{water solvent} - T_{solution}[/tex]
[tex]\Delta T_f = {0^{o}C} - ({-2.6^{o}C}) = 2.6^{o}C[/tex]
Substituting the values in equation (1):
[tex]\2.6^{o}C = i\times 1.86^{o}C/m\times 0.99 m[/tex]
[tex]i = \frac{2.6^{o}C}{1.86^{o}C/m\times 0.99 m} = 1.412[/tex]
Hence, the Van't Hoff factor (i) for [tex]HX[/tex] is 1.412.
