Search any question & find its solution
Question:
Answered & Verified by Expert
At S.T.P. the volume of $7.5 \mathrm{g}$ of a gas is $5.6 \mathrm{L}$. The gas is
Options:
Solution:
2433 Upvotes
Verified Answer
The correct answer is:
NO
Given, mass of gas $(W)=7.5 \mathrm{g}$
Volume of gas at $\operatorname{STP}(V)=56 \mathrm{L}$
$\because$ Moles of gas at $\mathrm{STP}=\frac{V(\mathrm{L})}{22.4(\mathrm{L})}=\frac{W}{M}$
Where $M=$ molar mass of gas.
$\therefore \quad M=\frac{W \times 22.4}{V}$
$$
=\frac{7.5 \times 22.4}{5.6}=30.00 \mathrm{g} \mathrm{mol}^{-1}
$$
Among the given options Molar mass $(M)$ of
$\mathrm{NO}=14+16=30.00 \mathrm{g} \mathrm{mol}^{-1}$
(b) $\mathrm{N}_{2} \mathrm{O}=28+16=44.00 \mathrm{g} \mathrm{mol}^{-1}$
(c) $\mathrm{CO}=12+16=28.00 \mathrm{g} \mathrm{mol}^{-1}$
(d) $\mathrm{CO}_{2}=12+32=44.00 \mathrm{g} \mathrm{mol}^{-1}$
$\because$ Molar mass of $\mathrm{NO}=30 \mathrm{g} \mathrm{mol}^{-1}$
Thus, the given gas is NO. Hence, is the correct option.
Volume of gas at $\operatorname{STP}(V)=56 \mathrm{L}$
$\because$ Moles of gas at $\mathrm{STP}=\frac{V(\mathrm{L})}{22.4(\mathrm{L})}=\frac{W}{M}$
Where $M=$ molar mass of gas.
$\therefore \quad M=\frac{W \times 22.4}{V}$
$$
=\frac{7.5 \times 22.4}{5.6}=30.00 \mathrm{g} \mathrm{mol}^{-1}
$$
Among the given options Molar mass $(M)$ of
$\mathrm{NO}=14+16=30.00 \mathrm{g} \mathrm{mol}^{-1}$
(b) $\mathrm{N}_{2} \mathrm{O}=28+16=44.00 \mathrm{g} \mathrm{mol}^{-1}$
(c) $\mathrm{CO}=12+16=28.00 \mathrm{g} \mathrm{mol}^{-1}$
(d) $\mathrm{CO}_{2}=12+32=44.00 \mathrm{g} \mathrm{mol}^{-1}$
$\because$ Molar mass of $\mathrm{NO}=30 \mathrm{g} \mathrm{mol}^{-1}$
Thus, the given gas is NO. Hence, is the correct option.
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.