Question

Mole fraction of solute in an aqua solution which boils at 100. 104 degree celsius for water is 0.52

Answer 1

The molality is calculated from the boiling point elevation

m=ΔtKbm=ΔtKb

=0.17∘C0.512∘C=0.17∘C0.512∘C = 0.33 m

The Freezing point depression is calculated from that molality

Δt=Ktm=(1.86∘C/m0.33m)=0.62∘CΔt=Ktm=(1.86∘C/m0.33m)=0.62∘C

The frezzing point is

t=0.00−0.62∘C=−0.62∘C