Total number of atoms present in 34 g of NH3 is

Question

Total number of atoms present in 34 g of NH3 is (A) 4×1023 (B) 4.8×1021 (C) 2×1023 (D) 48×1023

Tardigrade Admin · Accepted Answer

No. of moles of 34 g of NH3 = (34/17) = 2 No. of molecules = 2 × 6.023 × 1023 No. of atoms in one molecule of NH3 = 4 No. of atoms in 2 × 6.023 × 1023 molecules of NH3 = 4× 2 × 6.023 × 1023 = 48.18 × 1023

Q. Total number of atoms present in $34 g$ of $N H_{3}$ is

$4 \times 1 0^{23}$

$4.8 \times 1 0^{21}$

$2 \times 1 0^{23}$

$48 \times 1 0^{23}$

No. of moles of $34 g$ of $N H_{3} = \frac{34}{17} = 2$
No. of molecules $= 2 \times 6.023 \times 1 0^{23}$
No. of atoms in one molecule of $N H_{3} = 4$
No. of atoms in $2 \times 6.023 \times 1 0^{23}$ molecules of $N H_{3}$
$= 4 \times 2 \times 6.023 \times 1 0^{23}$
$= 48.18 \times 1 0^{23}$

Q. Total number of atoms present in 34g of NH3​ is

4×1023

4.8×1021

2×1023

48×1023