A 0.1215 g sample of solid magnesium (molar mass: 24.30 g $m o l^{- 1}$ ) is burned in bomb calorimeter (heat capacity = 1.7448 $k J^{0 C^{- 1}}$ ) which contain 300 mL of water. The temperature of water is raised by 1.0 °C, the energy of combustion of magnesium at 298 K is about

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$400 k J m o l^{- 1}$

$500 k J m o l^{- 1}$

$600 k J m o l^{- 1}$

$800 k J m o l^{- 1}$

answer is C.

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Detailed Solution

We have

$\begin{aligned} q_{water} = m_{water} c_{water} Δ T = (300 g) (4.184 {Jg}^{- 1}^{\circ} C^{- 1}) (1^{\circ} C) = 1255.2 J \\ q_{bomb} = (m_{bomb} c_{bomb}) Δ T = C_{bomb} Δ T = (1744.8 J^{\circ} C^{- 1}) (1^{\circ} C) = 1744.8 J \\ q_{toal} = q_{water} + q_{bomb} = (1255.2 + 1744.8) J = 3000 J \\ Δ_{c} U = \frac{q_{total}}{m / M} = \frac{3000 J}{(0.1215 g / 24.30 g {mol}^{- 1})} = 600 \times 10^{3} J {mol}^{- 1} \end{aligned}$