Answer to Question #176566 in Chemistry for [e-mail]

Question

Calculate the freezing and boiling point of a solution containing 5.57 g of sodium chloride (NaCl) and 65.0 g of water. (MM of NaCl = 58.44 g/mol, K_f of water = 1.86˚C/m, K_b of water = 0.52˚C/m)

Answer 1

Solution:

Calculate solution molality:

Molality of NaCl = Moles of NaCl / kilograms of water

m(NaCl) = (5.57 g NaCl) × (1 mol NaCl / 58.44 g NaCl) × (1 / 0.065 kg water) = 1.466 m NaCl

Molality of NaCl = 1.466 m

When NaCl dissolves, it separates into one Na⁺ ion and one Cl⁻ ion:

NaCl(aq) → Na⁺(aq) + Cl⁻(aq)

Because it breaks up into two ions, its van 't Hoff factor (i) is 2.

Thus:

BP - boiling point

∆T_BP = K_b × m × i

∆T_BP = (0.52˚C/m) × (1.466 m) × (2) = 1.52˚C

Water normally boils at 100˚C.

Hence,

BP = 100˚C + 1.52˚C = 101.52˚C

BP = 101.52˚C

FP - freezing point

∆T_FP = K_f × m × i

∆T_FP = (1.86˚C/m) × (1.466 m) × (2) = 5.45˚C

Water normally freezes at 0˚C.

Hence,

FP = 0˚C - 5.45˚C = -5.45˚C

FP = -5.45˚C

Answer:

The boiling point of a solution is 101.52˚C.

The freezing point of a solution is -5.45˚C.