Respuesta :
According the Graham's Law of effusion, if we let M₁ and R₁ be the mass and effusion rate of NH₃ and M₂ and R₂ be the mass and effusion rate of O₂, then:
R₂/R₁ = √(M₁/M₂).
Since M₁ = 14.01 + 3*1.01 = 17.04 and M₂ = 16 * 2 = 32, we see that:
R₂/R₁ = √(17.04/32) = 0.73
==> R₂ = 0.73 * R₁.
Thus, O₂ effuses at 73% the speed that NH₃ does
R₂/R₁ = √(M₁/M₂).
Since M₁ = 14.01 + 3*1.01 = 17.04 and M₂ = 16 * 2 = 32, we see that:
R₂/R₁ = √(17.04/32) = 0.73
==> R₂ = 0.73 * R₁.
Thus, O₂ effuses at 73% the speed that NH₃ does
Answer : The rate of effusion of ammonia gas is 1.37 times of rate of effusion of oxygen gas.
Solution :
According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.
[tex]R\propto \sqrt{\frac{1}{M}}[/tex]
or,
[tex]\frac{R_1}{R_2}=\sqrt{\frac{M_2}{M_1}}[/tex]
where,
[tex]R_1[/tex] = rate of effusion of ammonia gas
[tex]R_2[/tex] = rate of effusion of oxygen gas
[tex]M_1[/tex] = molar mass of ammonia gas = 17 g/mole
[tex]M_2[/tex] = molar mass of oxygen gas = 32 g/mole
Now put all the given values in the above formula, we get :
[tex]\frac{R_1}{R_2}=\sqrt{\frac{32g/mole}{17g/mole}}[/tex]
[tex]\frac{R_1}{R_2}=1.37[/tex]
[tex]R_1=1.37\times R_2[/tex]
Therefore, from this we conclude that, the rate of effusion of ammonia gas is 1.37 times of rate of effusion of oxygen gas.
