Search any question & find its solution
Question:
Answered & Verified by Expert
The volume of $2.8 \mathrm{~g}$ of $\mathrm{CO}$ at $27^{\circ} \mathrm{C}$ and 0.821 atm pressure is
$$
\left(R=0.08210 \text { L. atm K }{ }^{-1} \mathrm{~mol}^{-1}\right)
$$
Options:
$$
\left(R=0.08210 \text { L. atm K }{ }^{-1} \mathrm{~mol}^{-1}\right)
$$
Solution:
2042 Upvotes
Verified Answer
The correct answer is:
3 litres
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 $=\frac{30 \times 2.8}{28}=3 \mathrm{~L}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.