Question

18 g of a non-volatile solute is dissolved in 200 g of \[\ce{H2O}\] freezes at 272.07 K. Calculate the molecular mass of solute (K_f for water = 1.86 K kg mol⁻¹)

Answer 1

Formula: ΔT_f = K_f × m

Here, ΔT_f = T_pure − T_solution

Therefore, ΔT_f = 273.15 − 272.07 = 1.08 K

now substituting the value of ΔT_f in the formula

ΔT_f = K_f × m

1.08 = 1.86K kg/mol × m

`m = ("number of moles of solute")/("mass of solvent") xx 1000`

or `T_f = (K_f xx w_B xx 1000)/(M_B xx 200)`

`1.08 = (1.86 xx 18 xx 1000)/(1.08 xx 200)`

M_B = 155.0120 g/mol