Question

$x\, g$ of $MgSO _{4}(i=1.8)$ is in $2.5 \,L$ of solution has an osmotic pressure of $2.463\, atm$ at $27^{\circ} C$. What is the value of $x$ in $g$ ?

• A. 33.2
• B. 6.6
• C. 3.3
• D. 16.6

Answer 1

Answer: (D)

Given,

Mass of $MgSO _{4}(w)=x\, g$

vant Hoff factor $(i)=1.8$

Volume of solution $(V)=2.5\, L$

Osmotic pressure $(\pi)=2.463\, atm$

Temperature $(T)=27+273=300 \,K$

Molar mass of $MgSO _{4}=24+32+64$

$=120 \,g \,mol ^{-1}$

$\because \pi=i C R T=i \times \frac{w}{M} \times \frac{1}{V} \times R T$

$ \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 x=\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 \,g $