Question

$xg$ of $MgS{O}_4(i=1.8)$ is in $2.5L$ of solution has an osmotic pressure of $2.463atm$ 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 $MgS{O}_4(w)=xg$

vant Hoff factor $(i)=1.8$

Volume of solution $(V)=2.5L$

Osmotic pressure $(\pi )=2.463atm$

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

Molar mass of $MgS{O}_4=24+32+64$

$=120gmo{l}^{-1}$

$\because \pi =iCRT=i\times \frac{w}{M}\times \frac{1}{V}\times RT$

$\pi =2.463=\frac{i\times x\times RT}{M\times V}$

$=\frac{1.8\times x\times 8.314\times 300}{120\times 2.5}$

$\therefore x=\frac{2.463\times 120\times 2.5}{1.8\times 0.082\times 300}$

$x=\frac{738.9}{44.28}=16.68\approx 16.6g$