Question

Equimolal solution of nacl and bacl2 are prepared in water. Freezing point of nacl is found to be -2degree



c. What freezing point do you except for bacl2 solution

Answer 1

Explanation:

assume there is 1 L of water ....and as it is equimolar solution so no.of moles of NaCl and BaCl2 will be same ...

depression in freezing point = i X Kf X m 

i = no.of particles after dissociation...

as NaCl dissociate as -

NaCl ------> Na+ + Cl-

so i = 1 + 1 = 2

Kf for water = 1.86 degree C/m

and as water freezes at 0 degree C...but this NaCl solution is freezing at -2 degree C

so depression in freezing point = 2

putting the values...

2 = 2 X 1.86 X m

m = 2 / 3.72 = 0.538 moles/kg

now this would also be the value of m for BaCl2

BaCl2 -----> Ba2+ + 2Cl-

so i = 1 + 2 = 3

depression in freezing point = 3 X 1.86 X 0.538 = 3.002

so freezing point of BaCl2 solution = 0 - 3.002 = -3.002 degree C

just remember one thing for water we take density 1 g/ml ...so 1 L of water = 1 kg of water

and molality = no.of moles of solute/amount of solvent (water) in kg 

so as denominator is 1 ....so numerator will be the same for both NaCl and BaCl2 

thats why i have taken same molality for NaCl and BaCl2 solution

hope this helps