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Calculate the number of moles of the gas having volume 2.5 liter at $300 \mathrm{~K}$ and $4.5 \mathrm{~atm}$ ? $\left(\mathrm{R}=0.0821 \mathrm{~atm} \mathrm{dm}^3 \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)$
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The correct answer is:
$0.46$
$\begin{aligned} & \mathrm{PV}=\mathrm{nRT} \\ & 4.5 \times 2.5=\mathrm{n} \times 0.0821 \times 300 \\ & \mathrm{n}=0.56\end{aligned}$
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