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$A$ and $B$ are ideal gases. The molecular weights of $A$ and $B$ are in the ratio of $1: 4$. The pressure of a gas mixture containing equal weights of $A$ and $B$ is $P$ atm. What is the partial pressure (in atm) of $B$ in the mixture?
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Verified Answer
The correct answer is:
$\frac{P}{5}$
Mol. wt. ratio of $A$ and $B=1: 4$
$\therefore$ mole ratio of $A$ and $B$, if equal weight of $A$ and $B$ are taken $=4: 1$
$\begin{aligned}
\therefore \text { partial pressure of } B & =\frac{1}{(1+4)} \times P \\
& =\frac{P}{5}
\end{aligned}$
$\therefore$ mole ratio of $A$ and $B$, if equal weight of $A$ and $B$ are taken $=4: 1$
$\begin{aligned}
\therefore \text { partial pressure of } B & =\frac{1}{(1+4)} \times P \\
& =\frac{P}{5}
\end{aligned}$
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