25 mL of an aqueous solution of KCl was foun | vedclass

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Question

(A) The reaction is: $KCl + AgNO_3 \rightarrow AgCl + KNO_3$. At the equivalence point,the millimoles of $KCl$ equal the millimoles of $AgNO_3$: $n(KC

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) The reaction is: $KCl + AgNO_3 ightarrow AgCl + KNO_3$. At the equivalence point,the millimoles of $KCl$ equal the millimoles of $AgNO_3$: $n(KCl) = 20 \ mL 	imes 1 \ M = 20 \ mmol = 0.02 \ mol$. Mass of the solution $= 25 \ mL 	imes 1 \ g \ mL^{-1} = 25 \ g$. Mass of $KCl$ solute $= 0.02 \ mol 	imes 74.5 \ g \ mol^{-1} = 1.49 \ g$. Mass of solvent $= 25 \ g - 1.49 \ g = 23.51 \ g = 0.02351 \ kg$. Molality $(m) = \frac{0.02 \ mol}{0.02351 \ kg} \approx 0.8507 \ mol \ kg^{-1}$. Since $KCl$ undergoes $100 \%$ ionization,the van't Hoff factor $i = 2$. Depression in freezing point $\Delta T_f = i 	imes K_f 	imes m = 2 	imes 2.0 	imes 0.8507 = 3.4028 \ K$. The nearest integer is $3$.

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