2 m aqueous solution of an electrolyte X_3Y_ | vedclass

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(A) The dissociation of the electrolyte is given by: $X_3Y_{2(aq)} \rightarrow 3X^{2+}_{(aq)} + 2Y^{3-}_{(aq)}$.Here,the number of ions produced per f

Vedclass · Accepted Answer

$375.08$

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(A) The dissociation of the electrolyte is given by: $X_3Y_{2(aq)} ightarrow 3X^{2+}_{(aq)} + 2Y^{3-}_{(aq)}$.Here,the number of ions produced per formula unit is $n = 3 + 2 = 5$.The degree of ionization is $\alpha = 25\% = 0.25$.The van't Hoff factor $i$ is calculated as: $i = 1 + (n - 1)\alpha = 1 + (5 - 1) 	imes 0.25 = 1 + 4 	imes 0.25 = 1 + 1 = 2$.The elevation in boiling point is given by: $\Delta T_b = i 	imes K_b 	imes m$.Substituting the values: $\Delta T_b = 2 	imes 0.52 	imes 2 = 2.08 \ K$.The boiling point of the solution is $T_b = T_b^0 + \Delta T_b = 373.15 + 2.08 = 375.23 \ K$. (Approximating $T_b^0$ as $373 \ K$,we get $375.08 \ K$).

$2 \ m$ aqueous solution of an electrolyte $X_3Y_2$ is $25\%$ ionized. The boiling point of the solution is .......... $K.$ ($K_b$ for $H_2O = 0.52 \ K \ kg/mol$)

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