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An ideal gas $(\mathrm{X})$ present in a vessel of volume $\mathrm{V} \mathrm{L}$ exerted a pressure of $16.4 \mathrm{~atm}$ at $200 \mathrm{~K}$. What is its concentration in $\mathrm{mol} \mathrm{L}^{-1}$ ?
(Given $\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ )
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(Given $\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ )
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Verified Answer
The correct answer is:
1.00
$\begin{aligned} & \text { } \mathrm{PV}=\mathrm{nRT} \Rightarrow \mathrm{P}=\frac{\mathrm{nRT}}{\mathrm{V}}=\mathrm{CRT} \\ & \Rightarrow \mathrm{C}=\frac{\mathrm{P}}{\mathrm{RT}}=\frac{16.4}{0.082 \times 200}=1.00 \mathrm{M}\end{aligned}$
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