(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

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