This problem involves Graham's Law of Diffusion, which states that the rate of diffusion of a gas is…

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This problem involves Graham's Law of Diffusion, which states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. Since the rate is inversely proportion

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Here's the solution to the diffusion problem.

This problem involves Graham's Law of Diffusion, which states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. Since the rate is inversely proportional to the time taken for diffusion, we can write:

\frac{\text{t}_{\text{gas 2}}}{\text{t}_{\text{gas 1}}} = \sqrt{\frac{\text{M}_{\text{gas 1}}}{\text{M}_{\text{gas 2}}}}

where $\text{t}$ is the time taken for diffusion and $\text{M}$ is the molar mass.

Identify the given values.
- Time for gas X ( $\text{t}_X$ ) = 25.3 s
- Time for Oxygen ( $\text{t}_{\text{O}_2}$ ) = 20 s
- Relative atomic mass of Oxygen (O) = 16
Calculate the molar mass of oxygen gas ( $\text{O}_2$ ).
Oxygen exists as a diatomic molecule, $\text{O}_2$ .
$\text{M}_{\text{O}_2} = 2 \times \text{Relative atomic mass of O} = 2 \times 16 = 32 \text{ g/mol}$
Apply Graham's Law of Diffusion.
Let gas 1 be Oxygen ( $\text{O}_2$ ) and gas 2 be the unknown gas X.
$\frac{\text{t}_X}{\text{t}_{\text{O}_2}} = \sqrt{\frac{\text{M}_X}{\text{M}_{\text{O}_2}}}$
Substitute the known values into the equation:
$\frac{25.3 \text{ s}}{20 \text{ s}} = \sqrt{\frac{\text{M}_X}{32 \text{ g/mol}}}$
Solve for the molecular mass of gas X ( $\text{M}_X$ ).
First, calculate the ratio of the times:
$1.265 = \sqrt{\frac{\text{M}_X}{32}}$
Square both sides of the equation to remove the square root:
$(1.265)^2 = \frac{\text{M}_X}{32}$ $1.600225 = \frac{\text{M}_X}{32}$
Now, multiply by 32 to find $\text{M}_X$ :
$\text{M}_X = 1.600225 \times 32$ $\text{M}_X = 51.2072 \text{ g/mol}$
Rounding to three significant figures, the molecular mass of gas X is 51.2 g/mol.

The molecular mass of gas X is $\text{51.2 g/mol}$ .

This problem involves Graham's Law of Diffusion, which states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. Since the rate is inversely proportion

Answer

Step-by-step solution

Identify the given values.

Calculate the molar mass of oxygen gas ( $\text{O}_2$ ).

Apply Graham's Law of Diffusion.

Solve for the molecular mass of gas X ( $\text{M}_X$ ).

Try one yourself

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More Biology Questions

This problem involves Graham's Law of Diffusion, which states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. Since the rate is inversely proportion

Answer

Step-by-step solution

Identify the given values.

Calculate the molar mass of oxygen gas (O2\text{O}_2O2​).

Apply Graham's Law of Diffusion.

Solve for the molecular mass of gas X (MX\text{M}_XMX​).

Try one yourself

Still stuck on this one?

More Biology Questions

Calculate the molar mass of oxygen gas ( $\text{O}_2$ ).

Solve for the molecular mass of gas X ( $\text{M}_X$ ).