Freezing point of an aqueous solution is − 0.186 ∘ C. If the values of K b and K f of water are respectively 0.52 K kg mol − 1 and 1.86 K kg mol − 1 then find elevation in boiling point in terms o

Question

Freezing point of an aqueous solution is − 0.186 ∘ C. If the values of K b ​ and K f ​ of water are respectively 0.52 K kg mol − 1 and 1.86 K kg mol − 1 then find elevation in boiling point in terms of Kelvin for same solution.

TestHub · Accepted Answer

Given freezing point of the aqueous solution is $- 0.18 6^{\circ} C$ .

Depression in freezing point, $Δ T_{f} = 0 - (- 0.186) = 0.186 K$ .

We know that $Δ T_{f} = i \cdot K_{f} \cdot m$ and $Δ T_{b} = i \cdot K_{b} \cdot m$ .

Taking the ratio of the two equations:

$\frac{Δ T _{b}}{Δ T _{f}} = \frac{K _{b}}{K _{f}}$

Substituting the given values:

$Δ T_{b} = 0.186 \cdot \frac{0.52}{1.86}$

$Δ T_{b} = 0.1 \cdot 0.52 = 0.052 K$

Answer: $0.052$

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