The boiling point of water in a 0.1 m molal | vedclass

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(A) $(1)$ For $0.1 \ m$ $AgNO_3$ solution,the van't Hoff factor $i = 2$. The elevation in boiling point is $\Delta T_b = i 	imes K_b 	imes m = 2 	i

Vedclass · Accepted Answer

$100.1, 2.50$

Vedclass · Answer

(A) $(1)$ For $0.1 \ m$ $AgNO_3$ solution,the van't Hoff factor $i = 2$. The elevation in boiling point is $\Delta T_b = i 	imes K_b 	imes m = 2 	imes 0.5 	imes 0.1 = 0.1 \ ^{\circ}C$. Thus,the boiling point of solution $A$ is $x = 100 + 0.1 = 100.1 \ ^{\circ}C$. $(2)$ Upon mixing equal volumes of $0.1 \ m$ $AgNO_3$ and $0.1 \ m$ $BaCl_2$,the concentration of each solute becomes $0.05 \ m$. The reaction is $Ag^+_{(aq)} + Cl^-_{(aq)} ightarrow AgCl_{(s)}$. After precipitation,the remaining ions in solution $B$ are $NO_3^-$ $(0.05 \ m)$,$Ba^{2+}$ $(0.05 \ m)$,and $Cl^-$ $(0.1 - 0.05 = 0.05 \ m)$. The total molality of ions is $0.05 + 0.05 + 0.05 = 0.15 \ m$. The elevation in boiling point is $\Delta T_b = K_b 	imes \sum m_{ions} = 0.5 	imes 0.15 = 0.075 \ ^{\circ}C$. The boiling point of solution $B$ is $100.075 \ ^{\circ}C$. The difference is $100.1 - 100.075 = 0.025 \ ^{\circ}C = 2.5 	imes 10^{-2} \ ^{\circ}C$. Therefore,$|y| = 2.5$.

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