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The mass of oxygen gas occupying a volume of 11.2 litres at a temperature $27^{\circ} \mathrm{C}$ and a pressure of $760 \mathrm{~nm}$ of mercury in kilograms is $[$ Molecular weight of oxygen $=32]$
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Verified Answer
The correct answer is:
$0.01456$
At STP $(273 \mathrm{~K}$ and at $760 \mathrm{~mm}$ of $\mathrm{Hg})$ the mass of oxygen $=32 \mathrm{~g}$
From
$$
\frac{P_1 V_1}{m_1 T_1}=\frac{P_2 V_2}{m_1 T_2}
$$
$$
\begin{aligned}
& P_1=760 \mathrm{~mm}, V_1=2241 \text { (at STP) } \\
& T_1=273 \mathrm{~K} \\
& P_2=760 \mathrm{~mm} \text { (given), } V_2=11.2 \text { it (given) } \\
& T_2=27+273=300 \mathrm{~K} \\
& \therefore \frac{760 \times 224}{32 \mathrm{~g} \times 273}=\frac{760 \times 11.2}{m_2 \times 300} \\
& m_2=\frac{32 \times 273}{300 \times 2}=14.56 \mathrm{~g} \\
& =0.01456 \mathrm{~kg} \\
&
\end{aligned}
$$
From
$$
\frac{P_1 V_1}{m_1 T_1}=\frac{P_2 V_2}{m_1 T_2}
$$
$$
\begin{aligned}
& P_1=760 \mathrm{~mm}, V_1=2241 \text { (at STP) } \\
& T_1=273 \mathrm{~K} \\
& P_2=760 \mathrm{~mm} \text { (given), } V_2=11.2 \text { it (given) } \\
& T_2=27+273=300 \mathrm{~K} \\
& \therefore \frac{760 \times 224}{32 \mathrm{~g} \times 273}=\frac{760 \times 11.2}{m_2 \times 300} \\
& m_2=\frac{32 \times 273}{300 \times 2}=14.56 \mathrm{~g} \\
& =0.01456 \mathrm{~kg} \\
&
\end{aligned}
$$
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