Question

O2(g effuses at a rate that is ______ times that of xe(g under the same conditions.

Answer 1

Answer:

2.02 times

Explanation:

According to Graham's law, the rate of effusion of two gases can be expresed as:

$\frac{Rate_{1}}{Rate_{2}} =\sqrt{\frac{M_{2}}{M_{1}} }$

Where M are the atomic/molecular masses of the gases.

In this problem, the subscript 1 refers to O₂ gas (32 g/mol), and the subscript 2 refers to Xe gas (131.293 g/mol).

Putting the data in the equation gives us:

$\frac{RateO_{2}}{RateXe}=\sqrt{\frac{131.293}{32} } =2.02$

So O₂ effuses at a rate that is 2.02 times faster than the rate of Xe.

Answer 2

${{\text{O}}_{\text{2}}}$ (g) effuses at a rate that is $\boxed{{\text{2}}{\text{.02}}}$ times that of Xe (g) under the same conditions.

Further Explanation:

Graham’s law of effusion:

Effusion is the process by which molecules of gas travel through a small hole from high pressure to the low pressure. According to Graham’s law, the effusion rate of a gas is inversely proportional to the square root of the molar mass of gas.

The expression for Graham’s law is as follows:

$\boxed{\text{R}\propto \dfrac{1}{\sqrt{\mu}}}$

Here,

R is the rate of effusion of gas.

$\mu$ is the molar mass of gas.

Higher the molar mass of the gas, smaller will be the rate of effusion and vice-versa.

The rate of effusion of ${{\text{O}}_{\text{2}}}$ is expressed as follows:

${{\text{R}}_{{{\text{O}}_{\text{2}}}}} \propto \dfrac{1}{{\sqrt {{{{\mu}}_{{{\text{O}}_{\text{2}}}}}} }}$                   ......(1)

Here,

${{\text{R}}_{{{\text{O}}_{\text{2}}}}}$ is the rate of effusion of ${{\text{O}}_{\text{2}}}$.

${{{\mu }}_{{{\text{O}}_{\text{2}}$  is the molar mass of ${{\text{O}}_{\text{2}}}$.

The rate of effusion of Xe is expressed as follows:

${{\text{R}}_{{\text{Xe}}}} \propto \dfrac{1}{{\sqrt {{{\mu }}_{{\text{Xe}}}}} }}$

                                     ......(2)

Here,

${{\text{R}}_{{\text{Xe}}}}$ is the rate of effusion of Xe.

${{\mu }}_{{\text{Xe}}}}$ is the molar mass of Xe.

On dividing equation (1) by equation (2),

$\dfrac{{{{\text{R}}_{{{\text{O}}_{\text{2}}}}}}}{{{{\text{R}}_{{\text{Xe}}}}}}=\sqrt{\dfrac{{{{{\mu }}_{{\text{Xe}}}}}}{{{{\mu }}_{{{\text{O}}_{\text{2}}}}}}}}$                   ......(3)

Rearrange equation (3) to calculate  ${{\text{R}}_{{{\text{O}}_{\text{2}}}}}$.

${{\text{R}}_{{{\text{O}}_{\text{2}}}}}=\left( {\sqrt {\dfrac{{{{{\mu }}_{{\text{Xe}}}}}}{{{{{\mu }}_{{{\text{O}}_{\text{2}}}}}}}} } \right){{\text{R}}_{{\text{Xe}}}}$

                                                    ......(4)

The molar mass of ${{\text{O}}_{\text{2}}}$ is 32 g/mol.

The molar mass of Xe is 131.29 g/mol.

Substitute these values in equation (4).

$\begin{aligned}{{\text{R}}_{{\text{Ne}}}}&=\left( {\sqrt {\frac{{{\text{131}}{\text{.29}}}}{{{\text{32}}}}} } \right){{\text{R}}_{{\text{Xe}}}}\\&= \left( {\sqrt {4.1028} } \right){{\text{R}}_{{\text{Xe}}}}\\&=2.02{{\text{R}}_{{\text{Xe}}}}\\\end{aligned}$

Therefore the rate of effusion of ${{\mathbf{O}}_{\mathbf{2}}}$ is 2.02 times the rate of effusion of Xe.

Learn more:

1. Which statement is true for Boyle’s law: brainly.com/question/1158880

2. Calculation of volume of gas: brainly.com/question/3636135

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Ideal gas equation

Keywords: Effusion, rate of effusion, molar mass, O2, Xe, 2.02 times, Graham’s law, inversely proportional, square root.