Question

If the degree of ionization of $KCl$ in water at 290 K is 0.86. Calculate the mass of $KCl$ which must be made up to $1d{m}^3$ of the aqueous solution to the same osmotic pressure as the 4.0 per solution of glucose at that temperature.

• A. $4.45gm$
• B. $8.9gm$
• C. $10.32gm$
• D. $20.64gm$

Answer 1

The correct option is B $8.9gm$

Let W g og $KCl$ be added
For KCl, $n^{\prime }=2$, $i=[1+(n^{\prime }-1\left)\alpha \right]=[1+(2-1)\times 0.86]=1.86$
Since the two solutions are isotonic, the product $i\times C$ is constant.
Hence. $i\left(KCl\right)\times C\left(KCl\right)=i\left(glucose\right)\times C\left(glucose\right)$
$\frac{W}{74.6\times 1}\times 1.86=\frac{40}{1000\times 180}$
$W=\frac{74.6\times 40}{1000\times 180\times 1.86}=8.9g$
Hence, the mass of KCl which must be made up to $1d{m}^3$ of aqueous solution to the same osmotic pressure as the 4.0 per solution of glucose at that temperature is 8.9 g