Question

Assuming complete dissociation of the solute, how many grams of kno3 must be added to 275 ml of water to produce a solution that freezes at − 14. 5 ∘c ? the freezing point for pure water is 0. 0 ∘c and kf is equal to 1. 86 ∘c kg mol−1.

Answer 1

The mass of KNO₃ that is required to be added to 275 ml of water is; 108.3 g

Dissolution of a Solute

We are given;

K f =  1.86 °C.kg/m

Mass of solution = 275 g = 0.275 Kg

Freezing point of solution = −14.5 °C

Freezing point of pure water = 0.0 °C

We know that;

Δ T = Freezing point of solve nt - Freezing point of solut ion

Thus;

Δ T =  0°C - ( -14.5 °C)

Δ T = 14.5 °C

Also;

Δ T = K f * m * i

where;

K f is free zing con stant

m is m o l a l i t y

i is V a n t H o f f factor

Since KNO₃ produces two particles then; i = 2

Making the m the subject of the formula gives;

m = ΔT /(K × i)

m =  14.5 °C/(1.86 °C . kg/m o l × 2)

m = 3.898 m

Now;

M o l a l i t y = number of mo les/mass of solution in kilograms

Thus;

number of mo les of solute = m o l a l i t y × mass of solution in kilograms

number of mo les of solute= 3.898 m × 0.275 Kg

number of mo les of solute= 1.072 moles

Thus;

Number of mo les = mass/mo lar mass

M o lar mass of KNO₃ = 101 g/mol

Mass = Number of mo les ×  101 g/mol

Mass = 1.072 moles ×  101 g/mol

Mass of KNO₃ = 108.3 g

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