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Standard VIII

Chemistry

Boiling

If a compound...

Question

If a compound $A B$ dissociates to the extent of $75 %$ in an aqueous solution, the molality of the solution which shows a $1.82 K$ rise in the boiling point of the solution is
(Given: Molal elevation constant for water $= 0.52 {K kg mol}^{- 1}$ )

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Solution

$\begin{matrix} A B & \to & A^{+} & + & B^{-} 1 - α & α & α \end{matrix}$
Given that
$α = 3 / 4$
$N = 2$
$i = [1 + (2 - 1) α]$
The rise in boiling point is
$Δ T_{b} = i \times K_{b} \times m$
$\Rightarrow 1.82 = [1 + (2 - 1) 3 / 4] \times 0.52 \times m$
$\Rightarrow m = \frac{1.82}{0.52 \times \frac{7}{4}} = \frac{7.28}{3.64} = 2$
$\Rightarrow m = 2 mol/kg$

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If a compound AB dissociates to the extent of 75% in an aqueous solution, the molality of the solution which shows a 1.82 K rise in the boiling point of the solution is (Given: Molal elevation constant for water =0.52K kg mol−1)

AB→A++B−1−ααα Given that α=3/4 N=2 i=[1+(2−1)α] The rise in boiling point is ΔTb=i×Kb×m ⇒1.82=[1+(2−1)3/4]×0.52×m ⇒m=1.820.52×74=7.283.64=2 ⇒m=2 mol/kg

If a compound $A B$ dissociates to the extent of $75 %$ in an aqueous solution, the molality of the solution which shows a $1.82 K$ rise in the boiling point of the solution is
(Given: Molal elevation constant for water $= 0.52 {K kg mol}^{- 1}$ )