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Q.
$0.06$ mole of $KNO_3$ solid is added to $100 \, cm^3$ of water at $298 \, k$. The enthalpy of $KNO_{3aq}$ solution is $35.8\, kJ\,mol^{-1}$. After the solute is dissolved the temperature of the solution will be
Given: $0.06$ mole $kNO _{3}$ added to $100\, cm ^{3}$ of water at $298\,k$.
$\Delta H =35 \cdot 8 kJ / mol$
we know that formula of heat of solution,
$AH$ water $=$ mass water $\times S _{\text {water }} \times \Delta T$
Here, $\Delta H$ for $0.06$ mole $=2.148\, KJ$
mars of water $1 cm ^{3}=1 ml$
$100\, cm 3=100\, ml \approx 100\, g$
$\therefore 2148=100 \times 0.004184 kJ \times \Delta T$
$\Rightarrow \Delta T =5 \cdot 133$
It is an endothermic process as heat absorbed So temperature will decrease.
$\therefore $ final temp", $T =298-5 \cdot 133$
$T =292.867\, k$
$T \approx 293\, k$