Question

A solution containing 1 g of sodium chloride in 100 g of water freezes at 0.604°C. Calculate the degree of dissociation of sodium chloride. (Na = 23, Cl = 35.5, K_f for water = 1.87 K mol⁻¹)

Answer 1

Given: Mass of NaCl = 1 g

Mass of water = 100 g = 0.1 kg

Depression in freezing point (ΔT_f) = 0.604°C

K_f for water = 1.87 K kg mol⁻¹

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Molality of NaCl:

m = `(1  g)/(58.5  g//mol xx 0.1  kg)`

= `1/58.5`

≈ 0.1709 mol/kg

ΔT_f = i K_f m

`i = (Delta T_f)/(K_f * m)`

= `0.604/(1.87 xx 0.1709)`

= `0.604/0.3196`

≈ 1.89

We know that,

i = 1 + α

1.89 = 1 + α

α = 1.89 − 1

α = 0.89

∴ Degree of dissociation (α) is 0.89.