JEE MAIN - Chemistry (2022 - 27th July Evening Shift - No. 16)
A gas (Molar mass = 280 $$\mathrm{~g} \mathrm{~mol}^{-1}$$) was burnt in excess $$\mathrm{O}_{2}$$ in a constant volume calorimeter and during combustion the temperature of calorimeter increased from $$298.0 \mathrm{~K}$$ to $$298.45$$ $$\mathrm{K}$$. If the heat capacity of calorimeter is $$2.5 \mathrm{~kJ} \mathrm{~K}^{-1}$$ and enthalpy of combustion of gas is $$9 \mathrm{~kJ} \mathrm{~mol}^{-1}$$ then amount of gas burnt is _____________ g. (Nearest Integer)
Answer
35
Explanation
$\Delta \mathrm{U}=\mathrm{C} \Delta \mathrm{T}$
$$ \begin{aligned} &=2.5 \times 10^{3} \times 0.45 \\\\ &=1.125 \mathrm{~kJ} \end{aligned} $$
Considering $\Delta \mathrm{H} \simeq \Delta \mathrm{U}$
$$ \Delta \mathrm{H}=9 \mathrm{~kJ} / \mathrm{mol} \simeq \Delta \mathrm{U} $$
$\therefore$ Mass of gas burnt $=\frac{1.125}{9} \times 280=35 \mathrm{~g}$
$$ \begin{aligned} &=2.5 \times 10^{3} \times 0.45 \\\\ &=1.125 \mathrm{~kJ} \end{aligned} $$
Considering $\Delta \mathrm{H} \simeq \Delta \mathrm{U}$
$$ \Delta \mathrm{H}=9 \mathrm{~kJ} / \mathrm{mol} \simeq \Delta \mathrm{U} $$
$\therefore$ Mass of gas burnt $=\frac{1.125}{9} \times 280=35 \mathrm{~g}$
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