Solutions Question 181
Question: The elevation in boiling point of a solution of 13.44 g/mol of $ CuCl_2 $ in 1 kg of water using the following information will be (Molecular weight of $ CuCl_2=134.4g/mol $ 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 ) $
BETA
