Question

A current of dry air was passed through a series of bulbs containing 1.25 g of a solute $A_{2}B$ in 50 g of water and then through pure water. The loss in mass of the former series of bulbs was 0.98 g and in the later series 0.01 g. If the molar mass of $A_{2}B$ is 80, the degree of dissociation of $A_{2}B$ is (divide answer by 10) :

Answer 1

$P^0$ is proportional to the loss in weight from pure water
$P_0-P$ is proportional to the loss in weight from solution
and $\frac{(P^0-P)}{P^0}=X$ where X is the mole fraction of solute
$\frac{0.98}{0.01}=X$
$X=0.0102$
But $X=\frac{n}{N}$
Here, n is the number of moles of solute and N is the number of moles of water. The number of moles is the ratio of the mass to molar mass
$N=\frac{50-1.25}{18}=2.708$
$n=XN=0.0102\times 2.708=0.02763$
If the solute is undissociated then the number of moles of solute will be
$\frac{1.25}{80}=0.01563$
The vant hoff's factor is the ratio of the number of moles of solute considering dissociation to the number of moles of solute for no dissociation. It is
$\frac{0.02763}{0.01563}=1.768$
Hence the degree of dissociation is $\alpha =\frac{i-1}{n-1}=\frac{1.768-1}{3-1}=\frac{0.768}{2}=0.385$
Here n is the number of ions obtained on dissociation of 1 molecule of solute.
Hence, the degree of dissociation is $38.5\%\approx 40\%$.