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Q.

$4 g$ ${NH}_{4} {NO}_{3}$ were dissolved in $100 g$ water in bomb calorimeter with heat capacity of calorimeter system $150 {JK}^{- 1}$ , The temperature dropped by $1.5 K .$ Enthalpy of solution of ${NH}_{4} {NO}_{3}$

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a

$4.5 KJ {mol}^{- 1}$

b

$- 4.5 KJ {mol}^{- 1}$

c

$- 450 KJ {mol}^{- 1}$

d

$450 KJ {mol}^{- 1}$

answer is C.

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Detailed Solution

The enthalpy of solution is found as,

$\frac{150 \times 1.5}{4} \times \frac{80}{1000} kJ / mole$

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