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This is the second of our many quizzes on Diffusion and effusion. Click on this to access more such competitive quizzes on Physical Chemistry.

Diffusion and effusion Quiz 2

chemistry

 When we are given two gases, we can relate the rate of diffusion or effusion with the densities, so find the correct formula:

 The rate of diffusion is directly proportional to the pressure of the gas and the rate of the diffusion is inversely proportional to the square root of the density of the gas.

$\text{We can relate the factors rate of diffusion or effusion, pressure,}\\ \text{and densities as: } \dfrac{{{r}_{1}}}{{{r}_{2}}}=\dfrac{{{P}_{1}}}{{{P}_{2}}}\sqrt{\dfrac{{{d}_{2}}}{{{d}_{1}}}}$ 

When we calculate the rate of diffusion or effusion, we use the formula: $\text{Rate of diffusion/effusion = }\dfrac{x\text{ of the diffused/effused gas}}{\text{time taken}}$. Then what is x in this formula:

   The term x is equal to the ratio of mass and density. 

$\text{The correct formula is:} \\ \text{Rate of diffusion/effusion = }\dfrac{\text{Volume of the diffused/effused gas}}{\text{time taken}} \\ \text{The x in the formula is Volume.}$

  Study the table given below and find the value of x: <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/04c686ab-45f9-4700-90e0-89f54262a047967041074916135567.png"/>

 We can use the formula:$\dfrac{{{r}_{1}}}{{{r}_{2}}}=\sqrt{\dfrac{{{T}_{1}}{{M}_{2}}}{{{T}_{2}}{{M}_{1}}}}$ 

 $\text{We can use the formula:} \\ \dfrac{{{r}_{1}}}{{{r}_{2}}}=\sqrt{\dfrac{{{T}_{1}}{{M}_{2}}}{{{T}_{2}}{{M}_{1}}}} \\ \text{Putting the values from the table into the formula, we can write:} \\ \dfrac{1.609}{1}=\sqrt{\dfrac{x\text{ x 44}}{55\text{ x 17}}} \\ \text{Squaring both the side, we get:} \\ \dfrac{2.588}{1}=\dfrac{x\text{ x 44}}{55\text{ x 17}} \\ x=\dfrac{2.588\text{ x 55 x 17}}{44} \\ x=\dfrac{2419.78}{44} \\ x={{54.9}^{\circ }}C$ 

The law of diffusion/effusion given by Graham has much importance, which of the following is incorrect?

 The formula given by Graham’s law is: $\dfrac{{{r}_{1}}}{{{r}_{2}}}=\sqrt{\dfrac{{{d}_{2}}}{{{d}_{1}}}}=\sqrt{\dfrac{{{M}_{2}}}{{{M}_{1}}}}$ 

$\text{From the given options, Graham's law of diffusion or effusion cannot be} \\ \text{used to determine the value of the gas constant.}$

$150$ ml of A gas diffuses at the same time as $100$ ml of $C{{O}_{2}}$, find the molecular mass of the gas.

 We can use the formula: $\dfrac{{{r}_{A}}}{{{r}_{C{{O}_{2}}}}}=\dfrac{{{v}_{A}}/t}{{{v}_{C{{O}_{2}}}}/t}=\sqrt{\dfrac{{{M}_{C{{O}_{2}}}}}{{{M}_{A}}}}$ $\dfrac{{{r}_{A}}}{{{r}_{C{{O}_{2}}}}}=\dfrac{{{v}_{A}}}{{{v}_{C{{O}_{2}}}}}=\sqrt{\dfrac{{{M}_{C{{O}_{2}}}}}{{{M}_{A}}}}$ 

 $\text{We can use the formula: } \\ \dfrac{{{r}_{A}}}{{{r}_{C{{O}_{2}}}}}=\dfrac{{{v}_{A}}/t}{{{v}_{C{{O}_{2}}}}/t}=\sqrt{\dfrac{{{M}_{C{{O}_{2}}}}}{{{M}_{A}}}} \\ \dfrac{{{r}_{A}}}{{{r}_{C{{O}_{2}}}}}=\dfrac{{{v}_{A}}}{{{v}_{C{{O}_{2}}}}}=\sqrt{\dfrac{{{M}_{C{{O}_{2}}}}}{{{M}_{A}}}} \\ \text{So, the final formula will be:} \\ \dfrac{{{v}_{A}}}{{{v}_{C{{O}_{2}}}}}=\sqrt{\dfrac{{{M}_{C{{O}_{2}}}}}{{{M}_{A}}}} \\ \text{Now putting the values in the formula above, we get:} \\ \dfrac{150}{100}=\sqrt{\dfrac{44}{x}} \\ \dfrac{3}{2}=\sqrt{\dfrac{44}{x}} \\ \text{Squaring the equation on both the sides, we get:} \\ \dfrac{9}{4}=\dfrac{44}{x} \\ x=\dfrac{44\text{ x 4}}{9}=19.55 \\ \text{The molecular mass of molecule A is 19.55.}$