Question

The number of atoms present in $4.25$ grams of $N{H}_3$ is approximately:

• A. $1\times {10}^{23}$
• B. $1.5\times {10}^{23}$
• C. $2\times {10}^{23}$
• D. $6.023\times {10}^{23}$

Answer 1

Answer: (D)

$6.023\times {10}^{23}$

No. of moles = Given mass is divided by molar mass.

No. of moles of ammonia = 4.25/17 = $0.25$ mol
No. of molecules of ammonia = $6.023\times {10}^{23}\times 0.25$

One molecule contains a total 4 atoms.
Thus, total no. of atoms in ammonia $=6.023\times {10}^{23}\times 0.25\times 4=6.023\times {10}^{23}$ atoms