Solutions Question 333

Question: The elevation in boiling point of a solution of 13.44 g of $ CuCl_2 $ in 1 kg of water using the following information will be (Molecular weight of $ CuCl_2=134.4g $ and $ K_{b}=0.52,Kkg,mo{l^{-1}} $ )

Options:

A) 0.16

B) 0.05

C) 0.1

D) 0.2

Show Answer

Answer:

Correct Answer: A

Solution:

(i) $ i=\frac{No.ofparticlesafterionisation}{No.ofparticlesbeforeionisation} $ (ii) $ \Delta T_{b}=i\times K_{b}\times m $

$ i=\frac{1+2\alpha }{1},=1+2\alpha $ Assuming 100% ionization So, $ i=1+2=3 $

$ \Delta 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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