Question

When 11.7 g of $NaCl$ are dissolved in 200 g of water the depression in freezing point is doubled than the depression caused by 342 g of cane sugar in 1000 g of water. From this information, find the degree of dissociation of NaCl particles.

Answer 1

$\because \Delta T=\frac{1000\times K_f\times w}{m\times W}$

For NaCl: $2\Delta T=\frac{1000\times K_f\times 11.7}{200\times m}$ .... (i)

For sugar: $\Delta T=\frac{1000\times K_f\times 342}{1000\times 342}$ .....(ii)

By eqs. (i) and (ii), $m_{NaCl}=29.25$ (This is experimental value)

Also,
$\underset{\underset{1-\alpha }{1}}{NaCl}\rightleftharpoons \underset{\underset{\alpha }{0}}{N{a}^{+}}+\underset{\underset{\alpha }{0}}{C{l}^{-}}$

Total mole at equilibrium $=1+\alpha$

$\therefore \frac{m_N}{m_{exp}}=1+\alpha$

or $1+\alpha =\frac{58.5}{29.25}$

or $\alpha =1$

Thus, NaCl is 100% ionised in solutions.