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Q. $x$ g of a non-electrolytic compound (molar mass = 200) is dissolved in $1.0$ litre of $0.05\, M \,NaCl$ solution. The osmotic pressure of this solution is found to be $4.92 \,atm$ at $27^{\circ}C$. Calculate the value of $‘x’$ Assume complete dissociation of $NaCl$ and ideal behaviour of this solution

Solutions

Solution:

(i)For NaCl $\pi=iCRT=2\times 0.05 \times 0.0821 \times 300=2.463\,atm$
(ii) For unknown compound,
$\pi=CRT=\frac{x}{200}\times 0.0821 \times300=0.1231\times atm$
Total osmotic pressure $\pi=\pi_{1}+\pi_{2}$
$4.92 =2.463+0.1231 \,X$
$X=19.959\,g$