The number of atoms in 4.25 g of NH3 is(a) 4 x 1023(b) 2 x 1023(c) 1 x

1.	The number of atoms in 4.25 g of NH3 is(a) 4 x 1023(b) 2 x 1023(c) 1 x 1023(d) 6 x 1023
Answer» The correct answer is (d) 6.023 × 10²³ One mole of a chemical entity can be defined as equal to 6.023×10²³particles of that substance, where “particles” can mean “atoms, molecules, and the different subatomic particles”. We shall calculate the moles of the compound given and thus, the number of particles. Formula Used: \(moles = \frac{mass}{molar\, mass}\) The molecular weight of ammonia is equal to the sum of the atomic weights of nitrogen and hydrogen atoms = [14+(3×1)]= 17 grams Hence, 17 grams of ammonia = 6.023×10²³molecules of ammonia. Therefore, 4.25g of ammonia \(= \frac{6.023 \times 10^{23}}{17} \times 4.25 = 1.505 \times 10^{23}\) molecules of ammonia. Now each molecule of ammonia has four atoms, one atom of nitrogen and three atoms of hydrogen. Hence, the total number of atoms in 1.505 x 10²³molecules of ammonia = 4×1.505 × 10²³ = 6.023 × 10²³atoms.

The number of atoms in 4.25 g of NH3 is(a) 4 x 1023(b) 2 x 1023(c) 1 x 1023(d) 6 x 1023

Answer»

The correct answer is (d) 6.023 × 10²³

One mole of a chemical entity can be defined as equal to 6.023×10²³particles of that substance, where “particles” can mean “atoms, molecules, and the different subatomic particles”. We shall calculate the moles of the compound given and thus, the number of particles.

Formula Used: \(moles = \frac{mass}{molar\, mass}\)

The molecular weight of ammonia is equal to the sum of the atomic weights of nitrogen and hydrogen atoms = [14+(3×1)]= 17 grams

Hence, 17 grams of ammonia = 6.023×10²³molecules of ammonia.

Therefore, 4.25g of ammonia \(= \frac{6.023 \times 10^{23}}{17} \times 4.25 = 1.505 \times 10^{23}\) molecules of ammonia.

Now each molecule of ammonia has four atoms, one atom of nitrogen and three atoms of hydrogen. Hence, the total number of atoms in 1.505 x 10²³molecules of ammonia

The number of atoms in 4.25 g of NH3 is(a) 4 x 1023(b) 2 x 1023(c) 1 x 1023(d) 6 x 1023

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment