Question

Total number of atoms present in $23\text{g}$ of $NaN{O}_2$ is:
$\text{Given: Atomic mass:}Na=23u,N=14u,O=16u{N}_A=6.023\times {10}^{23}$

• A. $6.023\times {10}^{23}$
• B. $2.01\times {10}^{23}$
• C. $4.02\times {10}^{23}$
• D. $8.04\times {10}^{23}$

Answer 1

The correct option is D $8.04\times {10}^{23}$

Molar mass of $NaN{O}_2=(23+14+32)\text{g}=69\text{g}$
Number of moles of $NaN{O}_2$ present in $23\text{g}=\frac{23}{69}=\frac{1}{3}$
Number of molecules of $NaN{O}_3$ in $\frac{1}{3}$ moles
$=\frac{1}{3}\times 6.023\times {10}^{23}\approx 2.01\times {10}^{23}$
$\because$ One $NaN{O}_2$ molecule contains $4$ atoms
$\therefore$ total number of atoms in $2.01\times {10}^{23}$ molecules $=4\times 2.01\times {10}^{23}=8.04\times {10}^{23}$