Question

0.06 mole of ${KNO}_3$ solid is added to $100{cm}^3$ of water at 298K. The enthalpy of ${KNO}_3$ aq solution is $35.8kJ{mol}^{-1}$. After the solute is dissolved the temperature of the solutions will be?

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

Answer 1

The correct option is B 293 K

Given:0.06mole $kN{O}_3$ added to $100{cm}^3$ 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, $△H$ for 0.06 mole $=2.148KJ$

mars of water $1{cm}^3=1ml$

$\begin{array}{c}100cm3=100ml\approx 100g\\ \therefore 2148=100\times 0.004184kJ\times \Delta T\end{array}$

$\Rightarrow \Delta T=5\cdot 133$

It is an endothermic process as heat absorbed

So temperature will decrease.

$\begin{array}{cc}\therefore \text{final temp",}T& =298-5\cdot 133\\ T& =292.867k\\ T& \approx 293k\end{array}$

so, covrect answer is $B$