Question

use your knowledge of freezing point depression to determine an approximate mass of sodium chloride needed to prepare a mixture containing 850.0 grams of water at -10.0 c?

Answer 1

we know that

depression in freezing point is given by

dTf = i x kf x m

also

dTf = freezing of pure solvent - freezing point of solution

also

freezing point of pure solvent (water in this case) = 0

so

dTf = 0 - (-10)

dTf = 10

now

NaCl ---> Na+ + Cl-

there are two particles in the solution

so

i = vanthoffs factor = 2

now

Kf for water is 1.86

so

using this information

we get

dTf = i x kf x m

10 = 2 x 1.86 x m

m = 2.688

now

molality = moles of NaCl x 1000 / mass of water (g)

so

2.688 = moles of NaClx 1000 / 850

moles of NaCl = 2.285

now

mass = moles x molar mass

so

mass of NaCl = 2.285 x 58.44

mass of NaCl = 133.53

so

133.53 grams of NaCl is required