Question

$5.6$ L of oxygen gas at STP contains :

• A. $6.02\times {10}^{23}$ atoms
• B. $3.01\times {10}^{23}$ atoms
• C. $1.505\times {10}^{23}$ atoms
• D. $0.7525\times {10}^{23}$ atoms

Answer 1

The correct option is B $3.01\times {10}^{23}$ atoms

Number of moles of oxygen gas in $5.6L$ can be calculated as,

Moles of $O_2=\frac{5.6}{22.4}=\frac{1}{4}mol$
Number of molecules of $O_2=$ $\frac{1}{4}\times N_A$
As, one molecule of oxygen gas ($O_2$) contains 2 atoms of oxygen($O$), So total number of atoms $=2\times \frac{1}{4}\times N_A$

Number of oxygen atoms = $\frac{6.02\times {10}^{23}}{2}=3.01\times {10}^{23}$

Option $B$ is correct.