Question

What is the freezing point of a solution made 1.31 mol of CHCl3 in 530.0 g of CCl4 (Kf = 29.8 °C/m, Tf = -22.9 °C)?

Answer 1

let freezing point of solution is To heezing point of ccly = To = -22.9°C Depression in freezing point = AT = TEO-Tf ST= - 22.9°C - Tf Kf = 29: 8 'C/m moles of CHCl3 = 1.31 mol mass of ccly = 530.09 = 0.530 kg (0, 1000g = 1kg) molality of solution = m = moles of Prela man of ccly m= 1:31 mol 2.47m 0.530 kg i van't Hoff factor is I for CHCl3 using point - colligative property, depression in freezing ATE ĮKEM - 2.29°C - Tf = 1x 2918 'C/m X 2,47 m TE= - 75.9°C Hence, the freezing point of solution is -75.9°C Please Rate

Answer 2

Please sheck your math: it's -22.9 in the very last part before the answer, not -2.29. The answer should be -96.6