Question

How many moles of dinitrogen tetroxide (N2O4) are in 6.49 x 109 particles?

Answer 1

Answer: There are $1.08 \times 10^{-14}$ moles $N_{2}O_{4}$ present in $6.49 \times 10^9$ particles.

Explanation:

According to the mole concept, 1 mole of a substance contains $6.022 \times 10^{23}$ particles.

Hence, number of moles present in $6.49 \times 10^9$ particles are calculated as follows.

No. of moles = $\frac{6.49 \times 10^9}{6.022 \times 10^{23}}$ mol

                      = $1.08 \times 10^{-14}$ mol

Thus, there are $1.08 \times 10^{-14}$ moles $N_{2}O_{4}$ present in $6.49 \times 10^9$ particles.

Answer 2

Answer:

1.08 × 10⁻¹⁴ moles of N2O4

Explanation:

Given data:

Moles of dinitrogen tetroxide = ?

Number of particles = 6.49×10⁹

Solution:

Avogadro number:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions , particles and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.  The number 6.022 × 10²³ is called Avogadro number.

1 mole contain 6.022 × 10²³ particles.

6.49×10⁹ particles × 1 mole / 6.022 × 10²³ particles

1.08 × 10⁻¹⁴ mol