$x \mathrm{~g}^{\circ} \mathrm{MgSO}_4(i=1.8)$ is in $2.5 \mathrm{~L}$ of solution has an osmotic pressure…
$x \mathrm{~g}^{\circ} \mathrm{MgSO}_4(i=1.8)$ is in $2.5 \mathrm{~L}$ of solution has an osmotic pressure of $2.463 \mathrm{~atm}$ at $27^{\circ} \mathrm{C}$. What is the value of $x$ in $\mathrm{g}$ ?
33.2
6.6
3.3
16.6
Solution
Given,
Mass of $\mathrm{MgSO}_4(w)=x \mathrm{~g}$
vant Hoff factor $(i)=1.8$
Volume of solution $(V)=2.5 \mathrm{~L}$
Osmotic pressure $(\pi)=2.463 \mathrm{~atm}$
Temperature $(T)=27+273=300 \mathrm{~K}$
Molar mass of $\mathrm{MgSO}_4=24+32+64=120 \mathrm{~g} \mathrm{~mol}^{-1}$
$
\begin{aligned}
\because \quad \pi & =i C R T=i \times \frac{w}{M} \times \frac{1}{V} \times R T \\
\pi & =2.463=\frac{i \times x \times R T}{M \times V} \\
& =\frac{1.8 \times x \times 8.314 \times 300}{120 \times 2.5} \\
\therefore \quad\left(\because R=0.82 \mathrm{~L}-\mathrm{atm} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right) & \\
\therefore & =\frac{2.463 \times 120 \times 2.5}{1.8 \times 0.082 \times 300} \\
x & =\frac{738.9}{44.28}=16.68 \approx 16.6 \mathrm{~g}
\end{aligned}
$
Hence, option (d) is the correct answer