MHT CET · Chemistry · States of Matter
If the mixture of \(7 g\) of \(N_2\) and \(8 g\) of \(\mathrm{Ar}\) in a cylinder has the total pressure \(27 \mathrm{bar}\). What is the partial pressure of \(N_2\) ? (Atomic mass of \(N=14 \mathrm{~g} \mathrm{~mol}^{-1}, \mathrm{Ar}=40 \mathrm{~g} \mathrm{~mol}^{-1}\) )
- A 18 bar
- B 12 bar
- C 15 bar
- D 9 bar
Answer & Solution
Correct Answer
(C) 15 bar
Step-by-step Solution
Detailed explanation
Moles of \(N_2=\frac{7}{28}=\frac{1}{4}\)
Moles of \(\mathrm{Ar}=\frac{8}{40}=\frac{1}{5}\)
Partial pressure \(=X_{N_2} \times P_T\)
Partial pressure of
\(\begin{aligned}
N_2 & =\frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{5}} \times 27 \\
& =\frac{\frac{1}{4}}{\frac{9}{20}} \times 27 \\
& =\frac{1}{4} \times \frac{20}{9} \times 27 \\
& =15 \mathrm{bar}
\end{aligned}\)
Moles of \(\mathrm{Ar}=\frac{8}{40}=\frac{1}{5}\)
Partial pressure \(=X_{N_2} \times P_T\)
Partial pressure of
\(\begin{aligned}
N_2 & =\frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{5}} \times 27 \\
& =\frac{\frac{1}{4}}{\frac{9}{20}} \times 27 \\
& =\frac{1}{4} \times \frac{20}{9} \times 27 \\
& =15 \mathrm{bar}
\end{aligned}\)
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