The elevation in boiling point of a solution of $13.44 g$ of $C u C l_{2}$ in 1 kg of water using the following information will be ( Molecular weight of $C u C l_{2} = 134.4$ and $K_{b} = 0.52 K m o l a l^{- 1}$

0.16

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0.05

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Solution

The correct option is A 0.16
$(i) i = \frac{No. of particles after ionisation}{No. of particles before ionisation} (ii) Δ T_{b} = i \times K_{b} \times m \begin{matrix} C u C l_{2} \to & C u^{2 +} + & 2 C l^{-} 1 & 0 & 0 (1 - α) & α & 2 α \end{matrix} i = \frac{1 + 2 α}{1} i == 1 + 2 α$
Assuming $100 %$ ionisation So, $i = 1 + 2 = 3$
$Δ T_{b} = 3 \times 0.52 \times 0.1 = 0.156 \approx 0.16 [m = \frac{13.44}{134.4} = 0.1]$

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Q. The elevation in boiling point of a solution of

13.44 g

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The elevation in boiling point of a solution of 13.44 g of CuCl2 in 1 kg of water using the following information will be ( Molecular weight of CuCl2=134.4 and Kb=0.52 K molal−1

The correct option is A 0.16(i) i=No. of particles after ionisationNo. of particles before ionisation(ii) ΔTb=i×Kb×mCuCl2→Cu2++2Cl−100(1−α)α2αi=1+2α1i==1+2α Assuming 100% ionisation So, i=1+2=3 ΔTb=3×0.52×0.1=0.156≈0.16 [m=13.44134.4=0.1]

The elevation in boiling point of a solution of $13.44 g$ of $C u C l_{2}$ in 1 kg of water using the following information will be ( Molecular weight of $C u C l_{2} = 134.4$ and $K_{b} = 0.52 K m o l a l^{- 1}$