A sample of dinitrogen tetraoxide (\text{N}_2\text{O}_4) gas occupies a volume of 5.6\text… — JAMB UTME

A sample of dinitrogen tetraoxide ( $\text{N}_2\text{O}_4$ ) gas occupies a volume of $5.6\text{ dm}^3$ at standard temperature and pressure (STP). What is the total number of oxygen atoms contained in this sample? $[\text{Molar volume of gas at STP} = 22.4\text{ dm}^3\text{mol}^{-1}, N_A = 6.02 \times 10^{23}\text{ mol}^{-1}]$

$6.02 \times 10^{23}$ Answer
B
$5.62 \times 10^{23}$
C
$1.51 \times 10^{23}$
D
$9.03 \times 10^{23}$

Answer

6.02 \times 10^{23}

oxygen atoms

The sample contains

0.25\text{ mol}

\text{N}_2\text{O}_4

gas because

5.6\text{ dm}^3 / 22.4\text{ dm}^3\text{mol}^{-1} = 0.25\text{ mol}

. Since each molecule of

\text{N}_2\text{O}_4

contains 4 oxygen atoms, the total amount of oxygen atoms is

0.25 \times 4 = 1.0\text{ mol}

. Multiplying

1.0\text{ mol}

by Avogadro's constant (

6.02 \times 10^{23}\text{ mol}^{-1}

) yields

6.02 \times 10^{23}

oxygen atoms.

Step-by-Step Solution

Calculate the number of moles of dinitrogen tetraoxide (

\text{N}_2\text{O}_4

) gas at STP

n(\text{N}_2\text{O}_4) = \frac{\text{Volume}}{\text{Molar Volume}} = \frac{5.6\text{ dm}^3}{22.4\text{ dm}^3\text{mol}^{-1}} = 0.25\text{ mol}

At STP, one mole of any ideal gas occupies

22.4\text{ dm}^3

Determine the moles of oxygen atoms present in

0.25\text{ mol}

\text{N}_2\text{O}_4

n(\text{O}) = 0.25\text{ mol} \times 4 = 1.0\text{ mol}

of oxygen atoms

Each molecule of

\text{N}_2\text{O}_4

contains 4 oxygen atoms.

Compute the absolute number of oxygen atoms using Avogadro's constant

N(\text{O}) = n(\text{O}) \times N_A = 1.0\text{ mol} \times 6.02 \times 10^{23}\text{ mol}^{-1} = 6.02 \times 10^{23}

atoms

One mole of any substance contains

N_A

particles.

Key Concept

Molar gas volume at STP and Avogadro's constant stoichiometric conversions

Estimated Time:2m 0s

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