Question

$0.06$ mole of $KN{O}_3$ solid is added to $100c{m}^3$ of water at $298k$. The enthalpy of $KN{O}_{3aq}$ solution is $35.8kJmo{l}^{-1}$. After the solute is dissolved the temperature of the solution will be

• A. 293 k
• B. 298 k
• C. 301 k
• D. 304 k

Answer 1

Answer: (A)

Given: $0.06$ mole $kN{O}_3$ added to $100c{m}^3$ of water at $298k$.
$\Delta H=35\cdot 8kJ/mol$
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 $=2.148KJ$
mars of water $1c{m}^3=1ml$
$100cm3=100ml\approx 100g$
$\therefore 2148=100\times 0.004184kJ\times \Delta T$
$\Rightarrow \Delta T=5\cdot 133$
It is an endothermic process as heat absorbed So temperature will decrease.
$\therefore$ final temp", $T=298-5\cdot 133$
$T=292.867k$
$T\approx 293k$