Under identical conditions of temperature and pressure, a 100\text{ cm}^3 sample of nitrog… — JAMB UTME

Under identical conditions of temperature and pressure, a $100\text{ cm}^3$ sample of nitrogen(II) oxide ( $NO$ ) diffuses through a porous container in $20\text{ seconds}$ . If $100\text{ cm}^3$ of an unknown oxide of nitrogen diffuses through the exact same porous container in $35\text{ seconds}$ , what is the molecular formula of the unknown oxide of nitrogen? $[N = 14, O = 16]$

$N_2O_4$ Answer
B
$NO_2$
C
$N_2O$
D
$N_2O_5$

Answer

The molecular formula of the unknown oxide is

N_2O_4

The correct answer is derived using Graham's Law of Diffusion (

\frac{t_2}{t_1} = \sqrt{\frac{M_2}{M_1}}

). Given that nitrogen(II) oxide (

NO

) has a molar mass of

30\text{ g/mol}

and diffuses in

20\text{ s}

, an oxide diffusing in

35\text{ s}

must have a molar mass of

30 \times (35/20)^2 = 91.875 \approx 92\text{ g/mol}

. The formula matching this molar mass is dinitrogen tetroxide (

N_2O_4

Step-by-Step Solution

Calculate the molar mass of the reference gas, nitrogen(II) oxide (

NO

M_1(NO) = 14 + 16 = 30\text{ g/mol}

Graham's law requires the molar mass of the reference gas to determine the unknown gas mass.

Apply Graham's Law of Diffusion in terms of time taken for equal volumes of gas to diffuse.

\frac{t_2}{t_1} = \sqrt{\frac{M_2}{M_1}}

, where

t_1 = 20\text{ s}

t_2 = 35\text{ s}

, and

M_1 = 30\text{ g/mol}

Rate of diffusion is inversely proportional to diffusion time for fixed volume, so

\frac{\text{Rate}_1}{\text{Rate}_2} = \frac{t_2}{t_1} = \sqrt{\frac{M_2}{M_1}}

Solve for the unknown molar mass

M_2

\frac{35}{20} = 1.75 \implies (1.75)^2 = \frac{M_2}{30} \implies 3.0625 = \frac{M_2}{30} \implies M_2 = 91.875 \approx 92\text{ g/mol}

Squaring the time ratio isolates the molar mass ratio.

Identify the formula of the nitrogen oxide with a molar mass of

92\text{ g/mol}

N_2O_4

2(14) + 4(16) = 28 + 64 = 92\text{ g/mol}

Matching the calculated molar mass to the chemical formula of oxides of nitrogen.

Key Concept

Graham's Law of Diffusion applied to Oxides of Nitrogen

Estimated Time:2m 0s

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