Question

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

Answer 1

The rate of effusion of a gas is inversely proportional to the square root of the molecular weight of the species. Now there will be differences among isotopomers but neglecting these and taking the avg mol wt of N2 = 28 and Xe = 132; Rate(N2)/Rate(Xe) = sqrt (132/28) = 2.17

Answer 2

Ne (g) effuses at a rate that is $\boxed{{\text{2}}{\text{.6}}}$ 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 Ne is expressed as follows:

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

                                                 ......(1)

Here,

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

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

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),

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

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

${{\text{R}}_{{\text{Ne}}}}=\left( {\sqrt {\dfrac{{{{{\mu }}_{{\text{Xe}}}}}}{{{{{\mu }}_{{\text{Ne}}}}}}} } \right){{\text{R}}_{{\text{Xe}}}}$     ......(4)

The molar mass of Ne is 20.17 g/mol.

The molar mass of Xe is 131.29 g/mol.

Substitute these values in equation (4).

$\begin{aligned}{{\text{R}}_{{\text{Ne}}}}&= \left({\sqrt {\frac{{{\text{131}}{\text{.29}}}}{{{\text{20}}{\text{.17}}}}} } \right){{\text{R}}_{{\text{Xe}}}}\\&= \left( {\sqrt {6.50917} } \right){{\text{R}}_{{\text{Xe}}}}\\&= 2.5513{{\text{R}}_{{\text{Xe}}}}\\&\approx 2.6{{\text{R}}_{{\text{Xe}}}}\\\end{aligned}$

Therefore the rate of effusion of Ne is 2.6 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, Ne, Xe, 2.6 times, Graham’s law, inversely proportional, square root.