Question

The freezing point of a solution containing 5.85 g of NaCl in 100 g of water is −3.348°C. Calculate Van’t Hoff factor ‘i’ for this solution. What will be the experimental molecular weight of NaCl?
(K_f for water = 1.86 K kg mol⁻¹, at. wt. Na = 23, Cl = 35.5)

Answer 1

ΔT_f = 3.348°C

W_B = 5.85g

W_A = 100g

K_f = 1.86 K kg mol⁻¹

Observed mol. mass of NaCl (M_B) 

`Delta T_f = K_f * (W_B  xx 1000)/(W_A xx M_B)`

`Delta M_B = (K_f * W_B xx 1000)/(W_A xx Delta T_f)`

= `(1.86 xx 5.85 xx 1000)/(100 xx 3.348)`

= `(18.6 xx 5.85)/3.348`

= 32.5 g mol⁻¹

Normal mol. mass of NaCl = 23 + 35.5 = 58.5 g mol⁻¹

Van’t Hoff factor, i = `58.5/32.5` = 1.8

Answer 2

Given: Mass of NaCl = 5.85 g

Mass of water = 100 g = 0.1 kg

Freezing point depression (ΔT_f) = 3.348°C

K_f = 1.86 K kg mol⁻¹

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

Moles of NaCl = `5.85/58.5`

= 0.1 mol

Molality (m) = `0.1/0.1`

= 1 mol/kg

ΔT_f​ = i K_f​ m

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

= `3.348/(1.86 xx 1)`

= 1.8

We know that

`i = "Theoretical molar mass"/"Experimental molar mass"`

`"Experimental molar mass" = 58.5/1.8`

= 32.5 g/mol