Question

A sample of oxygen gas (o2) was found to effuse at a rate equal to three times that of an unknown gas. the molecular weight of the unknown gas is ________ g/mol.

Answer 1

Answer:
              288 g.mol⁻¹

Data Given:

Rate of Effusion of O₂  =  Rₐ  =  3

Rate of Effusion of Unknown  =  Rₓ  =  1

Molecular Mass of Oxygen  =  Mₐ  =  32 g.mol⁻¹

Molecular Mass of Unknown  =  Mₓ  =  ?

Formula Used,
                             Rₐ / Rₓ  =  $\sqrt{Mx/Ma}$

Putting Values,
                             3 / 1  =  $\sqrt{Mx/32}$
Or,
                             (3 / 1)²  =  ( $\sqrt{Mx/32}$ )²

                             9 / 1  =  Mx / 32

                             Mx  =  288 g.mol⁻¹

Answer 2

Graham's law of effusion states thta the rate of diffusion or effusion is inversely proportional to the square root of the molecular weight of compounds. The molecular weight of the unknown gas is 288 gmol⁻¹.

Given that,

• Rate of Effusion of O₂  =  Rₐ  =  3

• Rate of Effusion of Unknown  = Rₓ  =  1

• Molecular Mass of Oxygen  = Mₐ  =  32 gmol⁻¹

• Molecular Mass of Unknown  =  Mₓ  =  ?

Now, using the formula of rate of effusion:

• $\dfrac{\text {R}_a}{\text {R}_x} &= \sqrt \dfrac{\text{M}_x}{\text M_ a}$

Substituting the values:

• $\dfrac{3}{1} &= \sqrt \dfrac{\text{M}_x}{32}$
• $(\dfrac{3}{1})^2 &= \sqrt (\dfrac{\text{M}_x}{32})^2$
• $\dfrac{9}{1} &= \dfrac{\text{M}_x}{32}\\\\\text M_x&=288 \;\text {g/ mol}$

Therefore, the value of weight of unknown gas is equal to 288 g/mol.

To know more about Graham's law, refer to the following link:

https://brainly.com/question/12240656?referrer=searchResults