Question

Total number of atoms present in 34 g of $N{H}_3$ is:

• A. $4\times {10}^{23}$
• B. $4.8\times {10}^{21}$
• C. $2\times {10}^{23}$
• D. $48\times {10}^{23}$

Answer 1

The correct option is D $48\times {10}^{23}$

Mass of 1 mol or ($6.022\times {10}^{23}$ molecules) of $N{H}_3=14+3\times 1=17g$

Number o f molecules in 34 g $N{H}_3=\frac{34}{17}\times 6.022\times {10}^{23}=12.044\times {10}^{23}$

Number of atoms present in 1 molecule of $N{H}_3=4$

Number of atoms in 34 g $N{H}_3=4\times 12.044\times {10}^{23}$

$\approx 48\times {10}^{23}$

option D is correct