Search any question & find its solution
Question:
Answered & Verified by Expert
The maximum number of molecules is present in:
Options:
Solution:
2919 Upvotes
Verified Answer
The correct answer is:
$15 \mathrm{~L}$ of $\mathrm{H}_2$ gas at STP
At STP $22.4 \mathrm{~L} \mathrm{H}_2=6.023 \times 10^{23}$ molecules
$$
\begin{aligned}
15 \mathrm{~L} \mathrm{H}_2 & =\frac{6.023 \times 10^{23} \times 15}{22.4} \\
& =4.033 \times 10^{23} \\
5 \mathrm{~L} \mathrm{~N}_2 & =\frac{6.023 \times 10^{23} \times 5}{22.4} \\
& =1.344 \times 10^{23} \\
0.5 \mathrm{~g} \mathrm{H}_2 & =\frac{6.023 \times 10^{23} \times 0.5}{2} \\
& =1.505 \times 10^{23} \\
10 \mathrm{~g} \mathrm{of}_2 & =\frac{6.023 \times 10^{23} \times 10}{32} \\
& =1.882 \times 10^{23} .
\end{aligned}
$$
$$
\begin{aligned}
15 \mathrm{~L} \mathrm{H}_2 & =\frac{6.023 \times 10^{23} \times 15}{22.4} \\
& =4.033 \times 10^{23} \\
5 \mathrm{~L} \mathrm{~N}_2 & =\frac{6.023 \times 10^{23} \times 5}{22.4} \\
& =1.344 \times 10^{23} \\
0.5 \mathrm{~g} \mathrm{H}_2 & =\frac{6.023 \times 10^{23} \times 0.5}{2} \\
& =1.505 \times 10^{23} \\
10 \mathrm{~g} \mathrm{of}_2 & =\frac{6.023 \times 10^{23} \times 10}{32} \\
& =1.882 \times 10^{23} .
\end{aligned}
$$
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.