Question

Which of the following pair of solution(s) is/are isotonic?
(Assume all are 100% ionised)

• A. $0.2MBa(N{O}_3{)}_2$ and $0.3MN{a}_{2}S{O}_4$
• B. $0.15MNaCl$ and $0.1M{K}_{2}S{O}_4$
• C. $0.1MUrea$ and $0.05MNaCl$
• D. $0.2MBaC{l}_2$ and $0.2MUrea$

Answer 1

Answer: (B), (C)

$0.1MUrea$ and $0.05MNaCl$

Isotonic solutions are solutions having same osmotic pressure.
Osmotic pressure of a solution can be given by,
$\pi =iCRT$
$\pi$ = Osmotic pressure
C = Concentration ($mol/L$)
R = Gas constant ($Latmmo{l}^{-1}{K}^{-1}$)
T = Temperature ($K$)
$i$ = van't Hoff factor

(a)
$Ba(N{O}_3{)}_2\to B{a}^{2-}+2N{O}_3^{-}$
$\therefore$ $i=3$
${\pi }_{Ba(N{O}_3{)}_2}=3\times 0.2RT$
$N{a}_{2}S{O}_4\to 2N{a}^{+}+S{O}_4^{2-}$
$\therefore$ $i=3$
${\pi }_{N{a}_{2}S{O}_4}=3\times 0.3RT$
${\pi }_{N{a}_{2}S{O}_4}\ne {\pi }_{Ba(N{O}_3{)}_2}$
So, not isotonic

(b)
$NaCl\to N{a}^{+}+C{l}^{-}$
${\pi }_{NaCl}=2\times 0.15RT=0.3RT$
$K_{2}S{O}_4\to 2K^{+}+S{O}_4^{2-}$
$\therefore$ $i=3$
${\pi }_{K_{2}S{O}_4}=3\times 0.1RT=0.3RT$
Since
${\pi }_{K_{2}S{O}_4}={\pi }_{NaCl}$
So, isotonic

(c)
For urea :
Urea is a non-electrolyte so, $i=1$
${\pi }_{Urea}=0.1RT$
For $NaCl$:
${\pi }_{NaCl}=2\times 0.05RT=0.1RT$
${\pi }_{urea}={\pi }_{NaCl}$
So, isotonic

(d)
$BaC{l}_2\to B{a}^{2+}+2C{l}^{-}$
$\therefore$ $i=3$
${\pi }_{BaC{l}_2}=3\times 0.2RT$
Urea is a non-electrolyte so, $i=1$
${\pi }_{Urea}=0.2RT$
Since ,
${\pi }_{BaC{l}_2}\ne {\pi }_{NaCl}$
So, not isotonic