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A gaseous mixture was prepared by taking equal moles of $\mathrm{CO}$ and $\mathrm{N}_2$. If the total pressure of the mixture was found 1 atmosphere, the partial pressure of the nitrogen $\left(\mathrm{N}_2\right)$ in the mixture is
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The correct answer is:
$0.5 \mathrm{~atm}$
$$
\begin{array}{rlrl}
\because & { }^n \mathrm{CO} & ={ }^n \mathrm{~N}_2 \\
& \therefore & p_{\mathrm{CO}} & =p_{\mathrm{N}_2} \\
\text { Given, } & p_{\mathrm{CO}}+p_{\mathrm{N}_2} & =1 \mathrm{~atm} \\
& \text { or } & 2 p_{\mathrm{N}_2} & =1 \mathrm{~atm} \\
\text { or } & p_{\mathrm{N}_2} & =0.5 \mathrm{~atm}
\end{array}
$$
\begin{array}{rlrl}
\because & { }^n \mathrm{CO} & ={ }^n \mathrm{~N}_2 \\
& \therefore & p_{\mathrm{CO}} & =p_{\mathrm{N}_2} \\
\text { Given, } & p_{\mathrm{CO}}+p_{\mathrm{N}_2} & =1 \mathrm{~atm} \\
& \text { or } & 2 p_{\mathrm{N}_2} & =1 \mathrm{~atm} \\
\text { or } & p_{\mathrm{N}_2} & =0.5 \mathrm{~atm}
\end{array}
$$
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