Solutions Question 95
Question: The elevation in boiling point of a solution of 13.44 g of $ CuCl_2 $ , (molecular weight = 134.4, $ K_{b} $ = 0.52 K $ molalit{y^{-1}} $ in 1 kg water using the given information will be
Options:
A) 0.16
B) 0.05
C) 0.1
D) 0.2
Show Answer
Answer:
Correct Answer: A
Solution:
$ CuCl_2\to C{u^{2+}}+2C{I^{-}} $
| Initial | 1 | 0 | 0 |
| Final | $ (1-\alpha ) $ | $ \alpha $ | $ 2\alpha $ |
As we know that $ i=\frac{No.ofparticlesafterionisation}{No.ofparticlesbeforeionisation} $
$ \Delta T_{b}=i\times k_{b}\times m $
$ \therefore i=\frac{1+2\alpha }{1}\Rightarrow i=1+2\alpha $ Assuming 100% ionization $ i=1+2=3 $ Molality $ =\frac{w_{solute}}{M_{solute}+{M_{solvent(KG)}}}=\frac{13.44}{134.4\times 1}=0.1 $
$ \therefore \Delta T_{b}=3\times 0.52\times 0.1=0.156\approx 0.16 $
BETA
