What is the freezing point of solution containing 3g of a non-volatile solute in 20g of water. Freezing point of pure water is 273K, K_(f) of water = 1.86 Kkg/mol. Molar mass of solute is 300 g/mol. T^(@)-T=K_(f)m

What is the freezing point of solution containing 3g of a non-volatile

1.	What is the freezing point of solution containing 3g of a non-volatile solute in 20g of water. Freezing point of pure water is 273K, K_(f) of water = 1.86 Kkg/mol. Molar mass of solute is 300 g/mol. T^(@)-T=K_(f)m
Answer» Solution :`m=(W_(2))/(M_(2)W_(1))` `=(3)/(300xx20)xx1000` `T^(@)-T=(1.86xx3xx1000)/(300xx20)=0.93K` `T=273K-0.93K=272.07K` `THEREFORE" Freezing point of the solution = 272.07 K"`