Question

A solution containing 1.23 g of calcium nitrate in 10 g of water, boils at 100.975°C at 760 mm of Hg. Calculate the van’t Hoff factor for the salt at this concentration.

(K_b for water = 0.52 K kg mol⁻¹, mol. wt. of calcium nitrate = 164 g mol⁻¹)

Answer 1

Given: Mass of solute (W_B) = 1.23 g

Mass of solvent (W_A) = 10 g = 0.010 kg

Boiling point of solution (T_b) =100.975°C

Boiling point of pure water = 100.0°C

Elevation in boiling point (ΔT_b) = 100.975 − 100 = 0.975°C

K_b = 0.52 K kg mol⁻¹

Molar mass of calcium nitrate = 164 g/mol

Molality (m) = `(W_B xx 1000)/(M xx W_A)`

= `(1.23 xx 1000)/(164 xx 10)`

= `1230/1640`

= 0.75 mol/kg

ΔT_b ​= i K_b m

`i = (Delta T_b)/(K_b * m)`

= `0.975/(0.52 xx 0.75)`

= `0.975/0.39`

i = 2.5

Calcium nitrate (Ca(NO₃)₂) dissociates as

\[\ce{Ca(NO3)2 -> Ca^2+ + 2NO3-}\]

Expected i = 3

Calculated i = 2.5 implies partial dissociation