Question

The time taken for a certain volume of gas to diffuse through a small hole was 2 min. Under similar conditions an equal volume of oxygen took 5.65 minute to pass. The molecular mass of the gas is:
A. 32.0
B. 11.33
C. 4.0
D. 8.0

Answer 1

This numerical is based on Graham’s law of diffusion. Use the formula, $\dfrac{\dfrac{{{\text{V}}_{\text{A}}}}{{{\text{t}}_{\text{A}}}}}{\dfrac{{{\text{V}}_{\text{B}}}}{{{\text{t}}_{\text{B}}}}}=\dfrac{\sqrt{{{\text{M}}_{\text{B}}}}}{\sqrt{{{\text{M}}_{\text{A}}}}}$, to find the molecular mass of the unknown gas. See the conditions under which diffusion is taking place and apply the formula carefully by looking at the given terms. Find the molar mass of ${{\text{O}}_{2}}$ and apply the formula.

Complete step by step answer:
According to Graham’s diffusion law, the rate of diffusion of any gas is proportional to the molar mass of the gas. Rate is inversely proportional to square root of molecular mass of the gas molecules. Mathematically, it is written as $\text{r}\propto \dfrac{1}{\sqrt{\text{M}}}$.
When we deal in two gases to compare the rate of diffusion, the formula used is $\dfrac{{{\text{r}}_{\text{A}}}}{{{\text{r}}_{\text{B}}}}=\dfrac{\sqrt{{{\text{M}}_{\text{B}}}}}{\sqrt{{{\text{M}}_{\text{A}}}}}$, where ${{\text{r}}_{\text{A}}}$ and ${{\text{r}}_{\text{B}}}$ are the rates of diffusion of gas A and gas B respectively. ${{\text{M}}_{\text{A}}}$ and ${{\text{M}}_{\text{B}}}$ are the molecular masses of gas A and gas B.
Rate of any gas is measured as the amount of its volume diffused per unit time. The formula of Graham’s Diffusion law is modified as $\dfrac{\dfrac{{{\text{V}}_{\text{A}}}}{{{\text{t}}_{\text{A}}}}}{\dfrac{{{\text{V}}_{\text{B}}}}{{{\text{t}}_{\text{B}}}}}=\dfrac{\sqrt{{{\text{M}}_{\text{B}}}}}{\sqrt{{{\text{M}}_{\text{A}}}}}$, where ${{\text{V}}_{\text{A}}}$ and ${{\text{V}}_{\text{B}}}$ are the volumes of gas A and gas B, ${{\text{t}}_{\text{A}}}$ and ${{\text{t}}_{\text{B}}}$ are the times taken by the gases for diffusion and ${{\text{M}}_{\text{A}}}$ and ${{\text{M}}_{\text{B}}}$ are the molecular masses of gas A and gas B.
Let us solve this numerical step-by-step using the formula;
Step (1): Find the molecular mass of oxygen gas.
Oxygen gas has the formula ${{\text{O}}_{2}}$. The mass of dioxygen is twice the molar mass of atomic oxygen. The molecular mass of oxygen gas is $2\times 16$ or 32 grams.
Let the molecular mass of unknown gas be M.
Step (2): Write the given values.
We have ${{\text{V}}_{\text{A}}}={{\text{V}}_{\text{B}}}$, equal volumes of gases is mentioned.
Time taken by gas A to diffuse is 2 min. So, ${{\text{t}}_{\text{A}}}$= 2 min and time taken by gas B $\left( {{\text{O}}_{2}} \right)$ to diffuse is 5.65 min. So, ${{\text{t}}_{\text{B}}}$= 5.65 min.
The value of ${{\text{M}}_{\text{A}}}$ be M and ${{\text{M}}_{\text{B}}}$ be 32 grams.
Step (3): Apply the formula, $\dfrac{\dfrac{{{\text{V}}_{\text{A}}}}{{{\text{t}}_{\text{A}}}}}{\dfrac{{{\text{V}}_{\text{B}}}}{{{\text{t}}_{\text{B}}}}}=\dfrac{\sqrt{{{\text{M}}_{\text{B}}}}}{\sqrt{{{\text{M}}_{\text{A}}}}}$.
By substituting the values, we get $\dfrac{\dfrac{\text{V}}{2}}{\dfrac{\text{V}}{5.65}}=\dfrac{\sqrt{32}}{\sqrt{\text{M}}}$.
The expression can be modified as $\dfrac{5.65}{2}=\sqrt{\dfrac{32}{\text{M}}}$.
The value of M will be 4.009 grams or 4 grams.

The molecular mass of the unknown gas is 4 grams, which is option ‘c’.

Note: The formula of formula of Graham’s Diffusion law is modified when pressures of gas A and gas B are not equal. The formula will be $\dfrac{{{\text{r}}_{\text{A}}}}{{{\text{r}}_{\text{B}}}}=\dfrac{{{\text{P}}_{\text{A}}}\times \sqrt{{{\text{M}}_{\text{B}}}}}{{{\text{P}}_{\text{B}}}\times \sqrt{{{\text{M}}_{\text{A}}}}}$, where ${{\text{r}}_{\text{A}}}$ and ${{\text{r}}_{\text{B}}}$ are the rate of diffusion of gas A and gas B, ${{\text{P}}_{\text{A}}}$ and ${{\text{P}}_{\text{B}}}$ are the pressures of gas A and gas B and ${{\text{M}}_{\text{A}}}$ and ${{\text{M}}_{\text{B}}}$ are the molecular masses of gas A and gas B.