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

Under identical conditions of temperature and pressure, a $100\text{ cm}^3$ sample of an unknown gas $Q$ diffuses through a porous plug in $80\text{ seconds}$ , whereas an equal volume of oxygen gas ( $O_2$ ) diffuses through the same plug in $40\text{ seconds}$ . What is the relative molecular mass of gas $Q$ ?
( $O = 16.0$ )

128Answer
B
64
C
16
D
8

Answer

The relative molecular mass of gas

Q

is 128.

According to Graham's Law of diffusion, the time

t

required for a fixed volume of gas to diffuse is directly proportional to the square root of its molar mass

M

. Therefore,

\frac{t_Q}{t_{O2}} = \sqrt{\frac{M_Q}{M_{O2}}}

. Substituting

t_Q = 80\text{ s}

t_{O2} = 40\text{ s}

, and

M_{O2} = 32

gives

\frac{80}{40} = \sqrt{\frac{M_Q}{32}}

, so

2 = \sqrt{\frac{M_Q}{32}}

. Squaring both sides gives

4 = \frac{M_Q}{32}

, leading to

M_Q = 128

. Hence, the correct answer is 128.

Step-by-Step Solution

Calculate the molar mass of oxygen gas (

O_2

)

M_{O2} = 2 \times 16.0 = 32\text{ g/mol}

Oxygen exists as a diatomic gas, so its molecular mass is twice its atomic mass.

Set up Graham's Law relating diffusion time and molar mass for equal volumes

\frac{t_Q}{t_{O2}} = \sqrt{\frac{M_Q}{M_{O2}}}

The rate of diffusion is inversely proportional to the square root of molar mass, meaning diffusion time for a fixed volume is directly proportional to the square root of molar mass.

Substitute the given times (

t_Q = 80\text{ s}

t_{O2} = 40\text{ s}

) into the equation

\frac{80}{40} = \sqrt{\frac{M_Q}{32}} \implies 2 = \sqrt{\frac{M_Q}{32}}

Simplifying the time ratio yields a factor of 2.

Square both sides of the equation to solve for

M_Q

2^2 = \frac{M_Q}{32} \implies 4 = \frac{M_Q}{32} \implies M_Q = 4 \times 32 = 128

Squaring removes the radical, leaving a simple linear equation to calculate the unknown molar mass.

Key Concept

Graham's Law of Diffusion (Time-Molar Mass Relationship)

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic