Question

If a solution of mercuric cyanide containing 3 g per litre has an osmotic pressure $3.092\times {10}^{4}N{m}^{-2}$ at 298 K, the degree of dissociation of mercuric cyanide is (Atomic mass of $Hg=200.6gmo{l}^{-1})$ (write only integer part).

Answer 1

The expression for the osmotic pressure is $\Pi =iCRT$ .
Substitute values in the above expression.
$\frac{3.092\times {10}^{4}N/m^2}{1.01\times {10}^{5}atm/N/m^2}=i\times \frac{3g}{252.63g/mol}\times 0.0821L\cdot atm/mol/K\times 298K$
$i=1.05$
The degree of dissociation is $\alpha =\frac{i-1}{n^{\prime }-1}=\frac{1.05-1}{3-1}=0.025$
Hence, the degree of dissociation is $2.5\%$.