A certain mass of a substance when dissolved in 100g $C_6{H}_6$ lowers the freezing point by  ${1.28}^{0}C$. The same mass of solute dissolved in 100 g of water lowers the freezing point by ${1.40}^{0}C$. If the substance has normal molecular mass in benzene and is completely dissociated in water, into how many ions does it dissociate in water?$K_f$ for $H_{2}O$ and $C_6{H}_6$ are 1.86 and 5.12 K $mo{l}^{-}kg$ respectively.

$\Delta T_f=\frac{1000\times K_f\times w}{m\times w}$

For the solution in benzene using the data given

$1.28=\frac{1000\times 5.12\times w}{m_w\times 100}\mathrm{....1}$

For the solution in water in which solute dissociates

$1.40=\frac{1000\times 1.86\times w}{m_{\exp }\times 100}\mathrm{....2}$

Dividing eq. (ii) by (i)

$i=\frac{m_N}{m_{\exp }}=\frac{1.40}{1.28}\times \frac{5.12}{1.86}=3.01=3.0$

Now, suppose that formula of solute is 

$\begin{array}{l}{A}_x{B}_y\rightleftharpoons x{A}^{+}+y{B}^{-}\\ 100\\ \left(1-\alpha \right)x\alpha y\alpha \end{array}$

$i=1-\alpha +x\alpha +y\alpha$

$i=3and\alpha =1\left(Giventhat\alpha =1\right)$

No of ions given (x+y) = 3