1.24 g of $A X_{2}$ (molar mass $124 \mathrm{~g} \mathrm{~mol}^{-1}$ ) is dissolved in 1 kg of water

Question

1.24 g of $A X_{2}$ (molar mass $124 \mathrm{~g} \mathrm{~mol}^{-1}$ ) is dissolved in 1 kg of water to form a solution with boiling point of $100.0156^{\circ} \mathrm{C}$, while 25.4 g of $A Y_{2}$ (molar mass $250 \mathrm{~g} \mathrm{~mol}^{-1}$ ) in 2 kg of water constitutes a solution with a boiling point of $100.0260^{\circ} \mathrm{C}$.\ $K_{b}\left(\mathrm{H}_{2} \mathrm{O}ight)=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$\ Which of the following is correct?\

Professor PC Thomas Classes · Answer

(b) : $\Delta T_{b}\left(ight.$ For $\left.A X_{2}ight)$\ $=100.0156-100=0.0156^{\circ} \mathrm{C}$ 
$ \begin{aligned} & \Delta T_{b}\left(	ext { for } A Y_{2}ight)=100.0260-100 \ & =0.0260^{\circ} \mathrm{C} \ & \Delta T_{b}=i 	imes K_{b} 	imes m \ & m_{\left(A X_{2}ight)}=\frac{1.24}{124} 	imes \frac{1}{1}=0.01 \mathrm{~m} \ & m_{\left(A Y_{2}ight)}=\frac{25.4}{250} 	imes \frac{1}{2}=0.05 \mathrm{~m} \end{aligned} $ 
$i=3$ (for both $A X_{2}$ and $A Y_{2}$ when completely ionised)
 $ \begin{aligned} & \Delta T_{b}\left(A X_{2}ight)=3 	imes 0.52 	imes 0.01=0.0156^{\circ} \mathrm{C} \ & \Delta T_{b}\left(A Y_{2}ight)=3 	imes 0.52 	imes 0.05=0.078^{\circ} \mathrm{C} \end{aligned} $
Experimental $\Delta T_{b}$ is $\frac{1}{3} \mathrm{rd}$ of expected $\Delta T_{b}$ for $A Y_{2}$ that means $A Y_{2}$ remains unionised.\