Chemistry, 18.07.2019 18:30, lilytimpsonx

The rate of effusion of oxygen to an unknown gas is 1.49. what is the other gas?

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answered: rainyshea72

The unknown gas is $\boxed{{\text{chlorine}}}$ .

Further Explanation:

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

${\text{Rate of effusion}}\left({\text{R}}\right)\propto\frac{1}{{\sqrt {{\mu}}}}$ ......(1)

Where $\mu$ is the molar mass of gas.

The rate of effusion of oxygen is expressed as

${{\text{R}}_{{\text{oxygen}}}}\propto \frac{1}{{\sqrt {{{{\mu }}__{{\text{oxygen}}}}}}}$ ......(2)

And, the rate of effusion of other gas is expressed as

${{\text{R}}_{{\text{gas}}}}\propto\frac{1}{{\sqrt{{{{\mu }}__{{\text{gas}}}}}}}$ ......(3)

Since rate of effusion of oxygen is 1.49 times of rate of effusion of the unknown gas. Therefore,

${{\text{R}}_{{\text{oxygen}}}}=1.49{{\text{R}}_{{\text{gas}}}}$ ......(4)

Substitute the value of ${{\text{R}}_{{\text{white phosphorus}}}}$ and ${{\text{R}}_{{\text{Neon}}}}$ from equation (2) and equation (3) respectively in equation (4).

$\frac{1}{{\sqrt{{{{\mu }}_{_{{\text{oxygen}}}}}}}}= 1.49\frac{1}{{\sqrt{{{{\mu }}__{{\text{gas}}}}}}}$ ......(5)

Simplify equation (5) as follows:

$\begin{aligned}\frac{1}{{\sqrt{{{{\mu }}__{{\text{oxygen}}}}}}}&=1.49\frac{1}{{\sqrt{{{{\mu }}__{{\text{gas}}}}}}}\hfill\\\sqrt{{{{\mu }}__{{\text{gas}}}}}&=1.49\left({\sqrt {{{{\mu }}__{{\text{oxygen}}}}}}\right) \hfill\\{{{\mu }}__{{\text{gas}}}}&={\left({1.49\left({\sqrt {{{{\mu }}__{{\text{oxygen}}}}}} \right)}\right)^2}\hfill\\\end{gathered}$

Therefore the molar mass of white phosphorus is,

${{{\mu }}__{{\text{gas}}}}={\left({1.49\left({\sqrt{{{{\mu }}__{{\text{oxygen}}}}}}\right)} \right)^2}$ ......(6)

The molar mass of oxygen is 32 g/mol.

Substitute the value of molar mass of oxygen in the equation (6).

$\begin{aligned}{{{\mu }}__{{\text{gas}}}}&= {\left( {1.49\left({\sqrt {32\;{\text{g/mol}}} } \right)} \right)^2}\\&=71.0432\;{\text{g/mol}}\\\end{gathered}$

Therefore, the calculated molar mass of the unknown gas is 71.0432 g/mol and we know the gas with the molar mass of 71.0432 g/mol is chlorine. So the unknown gas is chlorine.

Learn more:

1. Which is the oxidation-reduction reaction:

2. Calculation of volume of gas:

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Ideal gas equation

Keywords: Effusion, rate of effusion, oxygen, unknown gas, chlorine, 1.49, molar mass, atomic mass, rate of effusion of oxygen, rate of effusion of gas, molar mass of oxygen, Graham’s law of effusion.

answered: Guest

While unproven, it uses facts to support part of a conclusion. it is still factual and has predictive power, however.

answered: Guest

Idon't hope someone you

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