The molal freezing point constant for water is 1.86 k.kgmole^-1. the freezing point of 0.1m nacl solution is

Question

Question from depression of freezing point .

∆Tb=T°f-Tf

Answer 1

∆T = imK

∆T = change in freezing point = ?

i = van't Hoff factor = 2 for NaCl (Na+ and Cl-)

m = molality = 0.1

K = freezing point constatn = 1.86

∆T = (2)(0.1)(1.86) = 0.372 degrees

New freezing point will be -0.372 degrees C since normal freezing point is 0ºC