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An aqueous dilute solution containing non-volatile solute boils at $100.52^{\circ} \mathrm{C}$. What is the molality of solution?
$\left(K_b=0.52 \mathrm{~kg} \mathrm{~mol}^{-1} \mathrm{~K}\right.$, boiling temperature of water $=100^{\circ} \mathrm{C}$ )
Options:
$\left(K_b=0.52 \mathrm{~kg} \mathrm{~mol}^{-1} \mathrm{~K}\right.$, boiling temperature of water $=100^{\circ} \mathrm{C}$ )
Solution:
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Verified Answer
The correct answer is:
$0.1 \mathrm{~m}$
$\Delta T_b=K_b \cdot m$
$$
T_b-T_b^{\circ}=K_b \cdot m
$$
where, $K_b=$ molal elevation constant $m=$ molality
$$
\begin{aligned}
(100.052-100.00) & =(0.52)(m) \\
0.052 & =0.52 \times m \\
m & =0.1
\end{aligned}
$$
$$
T_b-T_b^{\circ}=K_b \cdot m
$$
where, $K_b=$ molal elevation constant $m=$ molality
$$
\begin{aligned}
(100.052-100.00) & =(0.52)(m) \\
0.052 & =0.52 \times m \\
m & =0.1
\end{aligned}
$$
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