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$4 \mathrm{~g}$ of an ideal gas occupies $5.6035 \mathrm{~L}$ of volume at $546 \mathrm{~K}$ and 2 atmosphere pressure. What is its molecular weight?
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The correct answer is:
$16$
$\begin{aligned} & W=4 \mathrm{~g}, V=5.6035 \mathrm{~L}, T=546 \mathrm{~K}, P=2 \mathrm{~atm} \\ & \\ & \qquad \frac{M}{W}=\frac{R T}{P V} \quad\left(\because n=\frac{W}{M}\right) \\ & \frac{M}{4}=\frac{0.0821 \times 546}{2 \times 5.6035} \\ & \therefore \quad M=16\end{aligned}$
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