The ratio of lowering of vapour pressure of 0.1M aqueous solutions of KCl, CuSO4 and Na2SO4 is - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

A
1: 2:3
B
2: 2:3
C
1:1:1
D
3: 2:1

Solution:

0.1M KC $\to$ 0.2 moles of ion $(KCl \to K^{+} + {Cl}^{-})$

$0.1 M {CuSO}_{4} \to 0.2 moles of ion ({CuSO}_{4} \to {Cu}^{2 +} + {SO}_{4}^{2 -})$ $0.1 M \to 0.3 moles of ion ({Na}_{2} {SO}_{4} \to 2 {Na}^{+} + {SO}_{4}^{2 -})$ Ratio of lowering of vapour pressure depends upon no. of moles of ions.

=0.2:0.2:0.3

$\Rightarrow 2 : 2 : 3$

The ratio of lowering of vapour pressure of 0.1M aqueous solutions of $KCl, {CuSO}_{4} and {Na}_{2} {SO}_{4} is$

Solution:

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