0.2 molal aqueous solution of an electrolyte | vedclass

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(D) The electrolyte $A_3B$ dissociates as: $A_3B \rightarrow 3A^+ + B^{3-}$.Number of ions produced per formula unit,$n = 4$.Degree of dissociation,$\

Vedclass · Accepted Answer

$373.38$

Vedclass · Answer

(D) The electrolyte $A_3B$ dissociates as: $A_3B ightarrow 3A^+ + B^{3-}$.Number of ions produced per formula unit,$n = 4$.Degree of dissociation,$\alpha = 0.90$.The van't Hoff factor $i = 1 + \alpha(n - 1) = 1 + 0.90(4 - 1) = 1 + 0.90(3) = 1 + 2.7 = 3.7$.The elevation in boiling point is given by $\Delta T_b = i 	imes K_b 	imes m$.$\Delta T_b = 3.7 	imes 0.52 \ K \ Kg \ mol^{-1} 	imes 0.2 \ mol \ Kg^{-1} = 0.3848 \ K$.The boiling point of the solution is $T_b = T_b^0 + \Delta T_b = 373 \ K + 0.3848 \ K = 373.3848 \ K \approx 373.38 \ K$.

$0.2 \ molal$ aqueous solution of an electrolyte $A_3B$ is $90\%$ ionised. The boiling point of the solution at $1 \ atm$ is ........ $K.$ ($K_b$ for $H_2O = 0.52 \ K \ Kg \ mol^{-1}$ and $B.pt.$ of $H_2O = 373 \ K$)

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