Respuesta :
Answer is: the other gas is chlorine (Cl₂).
rate of effusion of oxygen : rate of effusion
of gas = 1.49 : 1.
rate of effusion of oxygen = 1/√M(O₂).
rate of effusion of oxygen = 1/√32.
rate of effusion of oxygen = 0.176.
rate of effusion of gas = 1/√M(gas).
rate of effusion of oxygen = rate of effusion of gas · 1.49.
1/√M(gas) = 0.176 1.49
1/√M(gas) = 0.118.
√M(gas) = 1 ÷ 0.118 = 8.46.
M(gas) = 8.46² = 71.5.
The unknown gas is [tex]\boxed{{\text{chlorine}}}[/tex].
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.
[tex]{\text{Rate of effusion}}\left({\text{R}}\right)\propto\frac{1}{{\sqrt {{\mu}}}}[/tex] ......(1)
Where [tex]\mu[/tex] is the molar mass of gas.
The rate of effusion of oxygen is expressed as
[tex]{{\text{R}}_{{\text{oxygen}}}}\propto \frac{1}{{\sqrt {{{{\mu }}__{{\text{oxygen}}}}}}}[/tex] ......(2)
And, the rate of effusion of other gas is expressed as
[tex]{{\text{R}}_{{\text{gas}}}}\propto\frac{1}{{\sqrt{{{{\mu }}__{{\text{gas}}}}}}}[/tex] ......(3)
Since rate of effusion of oxygen is 1.49 times of rate of effusion of the unknown gas. Therefore,
[tex]{{\text{R}}_{{\text{oxygen}}}}=1.49{{\text{R}}_{{\text{gas}}}}[/tex] ......(4)
Substitute the value of [tex]{{\text{R}}_{{\text{white phosphorus}}}}[/tex] and [tex]{{\text{R}}_{{\text{Neon}}}}[/tex] from equation (2) and equation (3) respectively in equation (4).
[tex]\frac{1}{{\sqrt{{{{\mu }}_{_{{\text{oxygen}}}}}}}}= 1.49\frac{1}{{\sqrt{{{{\mu }}__{{\text{gas}}}}}}}[/tex] ......(5)
Simplify equation (5) as follows:
[tex]\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}[/tex]
Therefore the molar mass of white phosphorus is,
[tex]{{{\mu }}__{{\text{gas}}}}={\left({1.49\left({\sqrt{{{{\mu }}__{{\text{oxygen}}}}}}\right)} \right)^2}[/tex] ......(6)
The molar mass of oxygen is 32 g/mol.
Substitute the value of molar mass of oxygen in the equation (6).
[tex]\begin{aligned}{{{\mu }}__{{\text{gas}}}}&= {\left( {1.49\left({\sqrt {32\;{\text{g/mol}}} } \right)} \right)^2}\\&=71.0432\;{\text{g/mol}}\\\end{gathered}[/tex]
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: https://brainly.com/question/2973661
2. Calculation of volume of gas: https://brainly.com/question/3636135
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.
