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The molality of an aqueous dilute solution containing non-volatile solute is $0.1 \mathrm{~m}$. What is the boiling temperature (in $\left.{ }^{\circ} \mathrm{C}\right)$ of solution? (Boiling point elevation constant, $K_b=0.52 \mathrm{~kg} \mathrm{~mol}^{-1} \mathrm{~K}$; boiling temperature of water $=100^{\circ} \mathrm{C}$ ).
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Verified Answer
The correct answer is:
100.052
Given, $K_b=0.52 \mathrm{~kg} \mathrm{~mol}^{-1} \mathrm{~K}$
Boiling point of water $=100^{\circ} \mathrm{C}$
$$
\begin{aligned}
m & =\frac{\Delta T_b}{K_b} \\
0.1 & =\frac{\Delta T_b}{0.52}
\end{aligned}
$$
$$
\begin{gathered}
\Delta T_b=0.052 \\
\text { Boiling point of solution }\left(T_b\right)=100+\Delta T_b \\
=100+0.052=100.052
\end{gathered}
$$
Boiling point of water $=100^{\circ} \mathrm{C}$
$$
\begin{aligned}
m & =\frac{\Delta T_b}{K_b} \\
0.1 & =\frac{\Delta T_b}{0.52}
\end{aligned}
$$
$$
\begin{gathered}
\Delta T_b=0.052 \\
\text { Boiling point of solution }\left(T_b\right)=100+\Delta T_b \\
=100+0.052=100.052
\end{gathered}
$$
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