the unconditional guru

Molality of the given solution = 0.1539m

∆T_f = 273.15 – 271 = 2.15 K

Therefore ∆T_f  = K_f .m

K_f = ∆T_f / m

K_f = 2.15/0.1539

 Now mass of the solute W₂ = 5g

Molar mass of the solute M₂ = 180g

Mass of the solution W₁ = 100g

 Now the mass of solvent = 95g

Using the following formula:-

=4.08 K

So the freezing point of the solution will be = 273.15 – 4.08

                                                                                                = 269.07 K