Question

A certain mass of a substance, when dissolved in 100 g $C_6{H}_6$, lowers the freezing point by ${1.28}^{o}C$. The same mass of solute dissolved in 100 g water, lowers the freezing point by ${1.40}^{o}C$. If the substance has normal molar mass in benzene and is completely ionised in water, how many ions does it dissociate in water? ($K_f$ for $H_{2}O$ and $C_6{H}_6$ are 1.86, 5.12 K $mo{l}^{-1}$ kg)

Answer 1

The expression for the depression in the freezing point is as given below.
$\Delta T=\frac{1000\times K_f\times w}{m\times W}$
For $C_6{H}_6:1.28=\frac{1000\times 5.12\times w}{m_N\times 100}$ ....(i)
For $H_{2}O:1.40=\frac{1000\times 1.80\times w}{m_{exp}\times 1000}$ ....(ii)
$(\because$ The solute behaves as normal in $C_6{H}_6$ and ionises in $H_{2}O$)
By eqs. (i) and (ii)
van't Hoff's factor (i) $=\frac{m_N}{m_{exp}}=\frac{1.40}{1.28}\times \frac{5.12}{1.86}=3.0$
Since, solute is 100% ionised and thus, $\alpha =1$
Let solute be $A_x{B}_y$ then,
$A_x{B}_y=x{A}^{+}+y{B}^{-}$
Mole before dissociation 1 0 0
Mole after dissociation $1-\alpha x\alpha y\alpha$
$\therefore \left(i\right)=1-\alpha +x\alpha +y\alpha =(x+y)(\because \alpha =1)$
or No. of ions given by solute in water $=3$
Thus, in water, the substance dissociates into 3 ions.