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Calculate the vapour density of a gas, $12.8 \mathrm{~g}$ of which occupies $10 \mathrm{~L}$ at a pressure of $750 \mathrm{~mm}$ at $27^{\circ} \mathrm{C}$.
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Verified Answer
Vapour density $\mathrm{(V.D.)}$ $=\frac{\mathrm{W}}{2} \times \frac{\mathrm{RT}}{\mathrm{PV}}$
Given,
$\begin{aligned}
&\mathrm{W}=12.8 \mathrm{~g} \\
&\mathrm{R}=0.082 \text { litre-atm degree }{ }^{-1} \text { mole }^{-1} \\
&\mathrm{~T}=273+27=300 \mathrm{~K} \\
&\mathrm{P}=\frac{750}{760} \mathrm{~atm} \\
&\mathrm{~V}=10 \mathrm{~L}
\end{aligned}$
Substituting these values, we get
$\text { V.D. }=\frac{12.8 \times 0.082 \times 300}{\frac{750}{760} \times 10 \times 2}=15.92.$
Given,
$\begin{aligned}
&\mathrm{W}=12.8 \mathrm{~g} \\
&\mathrm{R}=0.082 \text { litre-atm degree }{ }^{-1} \text { mole }^{-1} \\
&\mathrm{~T}=273+27=300 \mathrm{~K} \\
&\mathrm{P}=\frac{750}{760} \mathrm{~atm} \\
&\mathrm{~V}=10 \mathrm{~L}
\end{aligned}$
Substituting these values, we get
$\text { V.D. }=\frac{12.8 \times 0.082 \times 300}{\frac{750}{760} \times 10 \times 2}=15.92.$
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