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$0.06$ mole of $\mathrm{KNO}_{3}$ solid is added to $100 \mathrm{~cm}^{3}$ of water at $298 \mathrm{~K}$. The enthalpy of $\mathrm{KNO}_{3}(a q)$ solution is $35.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$. After the solute is dissolved the temperature of the solution will be
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$293 \mathrm{~K}$
$0.06 \mathrm{~mole} \mathrm{} \mathrm{KNO}_{3}$ added to $100 \mathrm{~cm}^{3}$ of water at $298 \mathrm{~K}$
$\Delta H=35.8 \mathrm{~kJ} / \mathrm{mol}$
$\Delta H=m \times C \times \Delta T$
$\left[\begin{array}{l}\Delta H \text { for } 0.06 \text { mole }=35.8 \times 0.06=2.148 \mathrm{~kJ} \\ \text { Mass }=100 \mathrm{~cm}^{3}=100 \mathrm{~g}\end{array}\right]$
$\therefore 2.148=100 \times 0.004184 \mathrm{~kJ} \times \Delta T$
$\Delta T=5.133 \mathrm{~K}$
It is an endothermic process as heat absorbed.
So, temperature will decrease.
$\therefore$ Final temperature
$$
\begin{aligned}
&T=298-5.133 \\
&T=292867 \mathrm{~K} \simeq 293 \mathrm{~K}
\end{aligned}
$$
$\Delta H=35.8 \mathrm{~kJ} / \mathrm{mol}$
$\Delta H=m \times C \times \Delta T$
$\left[\begin{array}{l}\Delta H \text { for } 0.06 \text { mole }=35.8 \times 0.06=2.148 \mathrm{~kJ} \\ \text { Mass }=100 \mathrm{~cm}^{3}=100 \mathrm{~g}\end{array}\right]$
$\therefore 2.148=100 \times 0.004184 \mathrm{~kJ} \times \Delta T$
$\Delta T=5.133 \mathrm{~K}$
It is an endothermic process as heat absorbed.
So, temperature will decrease.
$\therefore$ Final temperature
$$
\begin{aligned}
&T=298-5.133 \\
&T=292867 \mathrm{~K} \simeq 293 \mathrm{~K}
\end{aligned}
$$
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