JEE MAIN - Chemistry (2023 - 25th January Evening Shift - No. 15)
The number of pairs of the solutions having the same value of the osmotic pressure from the following is _________.
(Assume 100% ionization)
A. 0.500 $$\mathrm{M~C_2H_5OH~(aq)}$$ and 0.25 $$\mathrm{M~KBr~(aq)}$$
B. 0.100 $$\mathrm{M~K_4[Fe(CN)_6]~(aq)}$$ and 0.100 $$\mathrm{M~FeSO_4(NH_4)_2SO_4~(aq)}$$
C. 0.05 $$\mathrm{M~K_4[Fe(CN)_6]~(aq)}$$ and 0.25 $$\mathrm{M~NaCl~(aq)}$$
D. 0.15 $$\mathrm{M~NaCl~(aq)}$$ and 0.1 $$\mathrm{M~BaCl_2~(aq)}$$
E. 0.02 $$\mathrm{M~KCl.MgCl_2.6H_2O~(aq)}$$ and 0.05 $$\mathrm{M~KCl~(aq)}$$
Answer
4
Explanation
$\pi=i C R T$
The following pairs of solutions have same value of osmotic pressure
(A) $0.500 \mathrm{M} ~\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{aq}) \mathrm{i}=1$ and $0.25 \mathrm{M} ~\mathrm{KBr}(\mathrm{aq})$ $i=2$
$$ \begin{aligned} & \pi_1=0.5 \times 1 \times R T =0.5 R T \\\\ \mathrm{KBr} & \rightleftharpoons \mathrm{K}^{+}+\mathrm{Br}^{-} \\\\ & \pi_2=0.25 \times 2 \times R T=0.5 R T \end{aligned} $$
Thus, $\pi_1=\pi_2$
(B) $0.100 \mathrm{M} ~\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right](\mathrm{aq}) \mathrm{i}=5$ and $0.100 \mathrm{M}$ $\mathrm{FeSO}_{4}\left(\mathrm{NH}_{4}\right)_{2}(\mathrm{aq}) \mathrm{i}=5$
$$ \begin{array}{ll} \mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] \rightleftharpoons 4 \mathrm{~K}^{+}+\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-} (i=5) \\\\ \quad \pi_1=0.1 \times 5 \times R T=0.5 R T & \\\\ \mathrm{FeSO}_4 \cdot\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4 \rightleftharpoons \mathrm{Fe}^{2+}+2 \mathrm{NH}_4^{+}+2 \mathrm{SO}_4^{2-} & (i=5) \\\\ \pi_2=0.1 \times 5 \times R T=0.5 R T & \end{array} $$
Thus, $\pi_1=\pi_2$
C. 0.05 $$\mathrm{M~K_4[Fe(CN)_6]~(aq)}$$ and 0.25 $$\mathrm{M~NaCl~(aq)}$$
$$ \begin{gathered} {K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] \rightleftharpoons 4 {~K}^{+}+\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-} (i=5) \\\\ \pi_1=0.05 \times 5 \times R T=0.25 R T \\\\ {NaCl} \rightleftharpoons \mathrm{Na}^{+}+\mathrm{Cl}^{-} \quad(i=2) \\\\ \pi_2=0.25 \times 2 R T=0.5 R T \end{gathered} $$
Thus, $\pi_1 \neq \pi_2$
(D) $0.15 \mathrm{M}~ \mathrm{NaCl}(\mathrm{aq}) \mathrm{i}=2$ and $0.10 \mathrm{M}~ \mathrm{BaCl}_{2}$ (aq) $i=3$
$ \pi_1=0.15 \times 2 \times R T=0.3 R T$
$$ \begin{aligned} \mathrm{BaCl}_2 \rightleftharpoons \mathrm{Ba}^{2+}+2 \mathrm{Cl}^{-}(i=3) \\\\ \pi_2=0.1 \times 3 \times R T=0.3 R T \end{aligned} $$
Thus, $\pi_1=\pi_2$
(E) $0.02 \mathrm{M}~ \mathrm{KCl} . \mathrm{MgCl}_{2} .6 \mathrm{H}_{2} \mathrm{O}(\mathrm{aq}) \mathrm{i}=5$ and $0.05 \mathrm{M}$ $\mathrm{KCl}(\mathrm{aq}) \mathrm{i}=2$
$ \mathrm{KCl} \cdot \mathrm{MgCl}_2 \cdot 6 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{K}^{+}+3 \mathrm{Cl}^{-}+\mathrm{Mg}^{2+} \quad(i=5)$
$$ \begin{aligned} & \pi_1=5 \times 0.02 R T=0.1 R T \\\\ & 0.05 \mathrm{M} \mathrm{KCl}, \mathrm{KCl} \rightleftharpoons \mathrm{K}^{+}+\mathrm{Cl}^{-}(i=2) \\\\ & \pi_2=0.05 \times 2 R T=0.1 R T \end{aligned} $$
Thus, $\pi_1=\pi_2$
The following pairs of solutions have same value of osmotic pressure
(A) $0.500 \mathrm{M} ~\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{aq}) \mathrm{i}=1$ and $0.25 \mathrm{M} ~\mathrm{KBr}(\mathrm{aq})$ $i=2$
$$ \begin{aligned} & \pi_1=0.5 \times 1 \times R T =0.5 R T \\\\ \mathrm{KBr} & \rightleftharpoons \mathrm{K}^{+}+\mathrm{Br}^{-} \\\\ & \pi_2=0.25 \times 2 \times R T=0.5 R T \end{aligned} $$
Thus, $\pi_1=\pi_2$
(B) $0.100 \mathrm{M} ~\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right](\mathrm{aq}) \mathrm{i}=5$ and $0.100 \mathrm{M}$ $\mathrm{FeSO}_{4}\left(\mathrm{NH}_{4}\right)_{2}(\mathrm{aq}) \mathrm{i}=5$
$$ \begin{array}{ll} \mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] \rightleftharpoons 4 \mathrm{~K}^{+}+\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-} (i=5) \\\\ \quad \pi_1=0.1 \times 5 \times R T=0.5 R T & \\\\ \mathrm{FeSO}_4 \cdot\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4 \rightleftharpoons \mathrm{Fe}^{2+}+2 \mathrm{NH}_4^{+}+2 \mathrm{SO}_4^{2-} & (i=5) \\\\ \pi_2=0.1 \times 5 \times R T=0.5 R T & \end{array} $$
Thus, $\pi_1=\pi_2$
C. 0.05 $$\mathrm{M~K_4[Fe(CN)_6]~(aq)}$$ and 0.25 $$\mathrm{M~NaCl~(aq)}$$
$$ \begin{gathered} {K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] \rightleftharpoons 4 {~K}^{+}+\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-} (i=5) \\\\ \pi_1=0.05 \times 5 \times R T=0.25 R T \\\\ {NaCl} \rightleftharpoons \mathrm{Na}^{+}+\mathrm{Cl}^{-} \quad(i=2) \\\\ \pi_2=0.25 \times 2 R T=0.5 R T \end{gathered} $$
Thus, $\pi_1 \neq \pi_2$
(D) $0.15 \mathrm{M}~ \mathrm{NaCl}(\mathrm{aq}) \mathrm{i}=2$ and $0.10 \mathrm{M}~ \mathrm{BaCl}_{2}$ (aq) $i=3$
$ \pi_1=0.15 \times 2 \times R T=0.3 R T$
$$ \begin{aligned} \mathrm{BaCl}_2 \rightleftharpoons \mathrm{Ba}^{2+}+2 \mathrm{Cl}^{-}(i=3) \\\\ \pi_2=0.1 \times 3 \times R T=0.3 R T \end{aligned} $$
Thus, $\pi_1=\pi_2$
(E) $0.02 \mathrm{M}~ \mathrm{KCl} . \mathrm{MgCl}_{2} .6 \mathrm{H}_{2} \mathrm{O}(\mathrm{aq}) \mathrm{i}=5$ and $0.05 \mathrm{M}$ $\mathrm{KCl}(\mathrm{aq}) \mathrm{i}=2$
$ \mathrm{KCl} \cdot \mathrm{MgCl}_2 \cdot 6 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{K}^{+}+3 \mathrm{Cl}^{-}+\mathrm{Mg}^{2+} \quad(i=5)$
$$ \begin{aligned} & \pi_1=5 \times 0.02 R T=0.1 R T \\\\ & 0.05 \mathrm{M} \mathrm{KCl}, \mathrm{KCl} \rightleftharpoons \mathrm{K}^{+}+\mathrm{Cl}^{-}(i=2) \\\\ & \pi_2=0.05 \times 2 R T=0.1 R T \end{aligned} $$
Thus, $\pi_1=\pi_2$
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