Question

What amount of dinitrogen contains 3.6 × 10¹⁸ molecules?

Options

• 0.167 mg

• 0.280 mg

• 1.67 mg

• 2.80 mg

Answer 1

0.167 mg

Explanation:

6.022 × 10²³ molecules of dinitrogen ≡ 28 g of N₂

∴ 3.6 × 10¹⁸ molecules of dinitrogen

= `(3.6 xx 10^18 xx 28)/(6.022 xx 10^23)`

= 16.7 × 10⁻⁵ g

= 0.167 mg