KCET · Chemistry · States of Matter
The volume of \(2.8 \mathrm{~g}\) of \(\mathrm{CO}\) at \(27^{\circ} \mathrm{C}\) and 0.821 atm pressure is
\[
\left(R=0.08210 \text { L. atm K }{ }^{-1} \mathrm{~mol}^{-1}\right)
\]
- A 1.5 litres
- B 3 litres
- C 30 litres
- D 0.3 litres
Answer & Solution
Correct Answer
(B) 3 litres
Step-by-step Solution
Detailed explanation
Given, amount of \(\mathrm{CO}=2.8 \mathrm{~g}\)
Temperature \(=27^{\circ} \mathrm{C}\)
Pressure \(=0.821 \mathrm{~atm}\)
\(R=0.08210 \mathrm{~L} \mathrm{~atm} \mathrm{~K}{ }^{-1} \mathrm{~mol}^{-1}\)
We know that, \(p V=R T\)
\(\therefore \quad V=\frac{R T}{p}=\frac{0.0821 \times 300}{0.821}=30 \mathrm{~L}\)
\(\therefore 28 \mathrm{~g}\) of CO occupy volume \(30 \mathrm{~L}\)
\(\therefore 2.8 \mathrm{~g}\) of CO occupy volume \(=\frac{30 \times 2.8}{28}=3 \mathrm{~L}\)
Temperature \(=27^{\circ} \mathrm{C}\)
Pressure \(=0.821 \mathrm{~atm}\)
\(R=0.08210 \mathrm{~L} \mathrm{~atm} \mathrm{~K}{ }^{-1} \mathrm{~mol}^{-1}\)
We know that, \(p V=R T\)
\(\therefore \quad V=\frac{R T}{p}=\frac{0.0821 \times 300}{0.821}=30 \mathrm{~L}\)
\(\therefore 28 \mathrm{~g}\) of CO occupy volume \(30 \mathrm{~L}\)
\(\therefore 2.8 \mathrm{~g}\) of CO occupy volume \(=\frac{30 \times 2.8}{28}=3 \mathrm{~L}\)
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