Q3 / 100 · BITSAT 2024 (Memory Based Paper 3) · Chemistry
A decimolar solution potassium ferrocyanide is $50 \%$ dissociated at 300 K . The osmotic pressure of solution is $\left(R=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)$
- Correct answerA7.48 atm
- B4.99 atm
- C3.74 atm
- D6.23 atm
Solution
$\alpha=0.5$ (50\% dissociated)
The reaction involves is
Total moles at equilibrium $=1-\alpha+4 \alpha+\alpha$
$\text { Total }=1+4 \alpha$
$\therefore$ van't Hoff factor $(i)=\frac{1+4 \alpha}{1}=1+4 \times 0.5=3$
Osmotic pressure, $\pi=i C R T$
$\begin{aligned}
C & =0.1 \mathrm{M}=10^2 \mathrm{~mol} \mathrm{~m}^{-3} \\
\pi & =3 \times 10^2 \times 8.314 \times 300 \\
& =7.483 \times 10^5 \mathrm{Nm}^{-2}
\end{aligned}$
or
$\pi=7.48 \mathrm{~atm}$
Total moles at equilibrium $=1-\alpha+4 \alpha+\alpha$
$\text { Total }=1+4 \alpha$
$\therefore$ van't Hoff factor $(i)=\frac{1+4 \alpha}{1}=1+4 \times 0.5=3$
Osmotic pressure, $\pi=i C R T$
$\begin{aligned}
C & =0.1 \mathrm{M}=10^2 \mathrm{~mol} \mathrm{~m}^{-3} \\
\pi & =3 \times 10^2 \times 8.314 \times 300 \\
& =7.483 \times 10^5 \mathrm{Nm}^{-2}
\end{aligned}$
or
$\pi=7.48 \mathrm{~atm}$Hence, the correct answer is option: 'A'.