A certain mass of a substance when dissolved in 100 g C₆H₆ lowers the freezing point by 1.28°C. The same mass of solute dissolved in 100 g of water lowers the freezing point by 1.40°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 H₂O and C₆H₆ are 1.86 and 5.12 K mol^-1 kg respectively.

Moderate

Solution

${ΔT}_{f} = \frac{1000 \times K_{j}^{'} \times w}{m \times W}$
For the solution in benzene using the data given
$1.28 = \frac{1000 \times 5.12 \times w}{m_{N} \times 100} (m_{N} = normal mol. mass) \dots ( i)$
For the solution in water in which solute dissociates
$1 .40 = \frac{1000 \times 1 .86 \times w}{m_{exp.} \times 100} . . . (ii)$
dividing e.q. (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
$A_{x} B_{y} ⇌ {xA}^{+} + {yB}^{-}$
$\begin{array}{ccc} 1 & 0 & 0 \\ (1 - α) & x α & y α \end{array}$
$∴ i = 1 - α + xα + yα$

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