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Q. If $20 \,g$ of a solute was dissolved in $500 \,mL$ of water and osmotic pressure of the solution was found to be $600\, mm$ of $Hg$ at $15^{\circ} C$, then molecular weight of the solute is:

BHUBHU 2004

Solution:

Osmotic pressure is colligative property. Use the formula $\pi V=n R T$ to solve problem.
$\pi V=n R T$
or $ \pi=\frac{n}{V} R T=\frac{m R T}{M V}$
where $\pi=$ osmotic presure
$=600\, mm$ of $Hg $
$=600 / 760 \,atm$
$m=$ mass of solute $=20\, g$
$M=$ molecular mass of solute
$R=0.0821 \,L\, atm\, mol ^{-1}\, K ^{-1} $
$T=15^{\circ} C =15+273=288\, K$
$V=500 \,mL =0.5\, L $
$\frac{600}{760}=\frac{20 \times 0.0821 \times 288}{M \times 0.5}$
or $M=1200$
$\therefore $ molecular mass of solute is $1200$ .