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To 100 $cm^3$ water at 298 K, 0.06 mole of $KNO_3$ is added. The enthalpy of $KNO_3$(aq) solution is 35.8 kJ $mol^{-1}$. What will be the temperature of the solution, after the solute is dissolved?

(a) 293 K
(b) 297 K
(c) 303 K
(d) 345 K

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Answer: 293 K
Dissolution of $KNO_3$ is endothermic.
$\Delta H_{sol}$ = 35.8 kJ $mol^{-1}$
$\therefore $ Heat absorbed when 0.06 kJ mole of $KNO_3$ is dissolved = $35.8 \times 0.06$ kJ = 2148 J
Heat absorbed, $q=m\times c \times \Delta T$
$\therefore 2148 J= 100 cm^3 \times 4.184 J g^{-1}K^{-1}\times \Delta T \Rightarrow \Delta T= 5K$
$\therefore$ Temperature of the solution = 298 - 5 = 293 K
answered Feb 27, 2014 by mosymeow_1

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