Respuesta :
Answer:
The molar mass of the unknown gas is 17.3 g/mol. The molar mass matches that of ammonia (NH₃) the most (17 g/mol)
Explanation:
Let the unknown gas be gas 1
Let N₂O₄ gas be gas 2
Rate of effusion ∝ [1/√(Molar Mass)]
R ∝ [1/√(M)]
R = k/√(M) (where k is the constant of proportionality)₁₂
R₁ = k/√(M₁)
k = R₁√(M₁)
R₂ = k/√(M₂)
k = R₂√(M₂)
k = k
R₁√(M₁) = R₂√(M₂)
(R₁/R₂) = [√(M₂)/√(M₁)]
(R₁/R₂) = √(M₂/M₁)
R₁ = 2.3 R₂
M₁ = Molar Mass of unknown gas
M₂ = Molar Mass of N₂O₄ = 92.01 g/mol
(2.3R₂/R₂) = √(92.01/M₁)
2.3 = √(92.01/M₁)
92.01/M₁ = 2.3²
M₁ = 92.01/5.29
M₁ = 17.3 g/mol
The molar mass matches that of ammonia the most (17 g/mol)
The unknown gas in the system has been ammonia.
The rate of diffusion of the two gases has been proportional to the molar mass of the gases.
The ratio of the rate of two gases can be given as:
[tex]\rm \dfrac{RateA}{RateB}\;=\;\sqrt{\dfrac{Molar\;mass\[A}{Molar\;mass\;B} }[/tex]
The two gases can be given as:
Gas A = Nitrogen tetraoxide = 2.3x
Gas B = x
[tex]\rm \dfrac{2.3x}{x}\;=\;\sqrt{\dfrac{92.011}{m} }[/tex]
Mass of the unknown gas = 17.39 grams.
The mass has been equivalent to the mass of the Ammonia. Thus, the unknown gas in the system has been ammonia.
For more information about the diffusion of gas, refer to the link:
https://brainly.com/question/879602
