Question

The pressure in a bulb dropped from 2000 to 1500 mm Hg in 47 min when the contained oxygen leaked through a small hole. The bulb was then evacuated. A mixture of oxygen and another gas of molecular weight 79 in the molar ratio of 1:1 at a total pressure of 4000 mm of mercury was introduced. Find the molar ratio of the two gases remaining in the bulb after a period of 74 min.
A. $\dfrac{{{M}_{gas}}}{{{M}_{{{O}_{2}}}}}=0.666$
B. $\dfrac{{{M}_{gas}}}{{{M}_{{{O}_{2}}}}}=1.236$
C. $\dfrac{{{M}_{gas}}}{{{M}_{{{O}_{2}}}}}=2.472$
D. $\dfrac{{{M}_{gas}}}{{{M}_{{{O}_{2}}}}}=5.236$

Answer 1

Hint: There is a relationship between rate of the gas and molar molecular weight of the gases and it is called Graham’s law. The formula of Graham’s law of diffusion is as follows.
\[\dfrac{{{r}_{gas}}}{{{r}_{{{O}_{2}}}}}=\sqrt{\dfrac{{{M}_{{{O}_{2}}}}}{{{M}_{gas}}}}\]
Here ${{r}_{g}}$ = rate of flow of gas
${{r}_{{{O}_{2}}}}$ = rate of flow of oxygen
${{M}_{{{O}_{2}}}}$ = Molecular weight of the oxygen
${{M}_{gas}}$ = Molecular weight of the gas

Complete answer:
- In the question it is given that the total pressure of the mixture of the oxygen and other gas is in the molar ratio of 1:1 is 4000 mm of Hg, means pressure of the each gas exerted is 2000 mm of Hg.
- The change in pressure in 47 min = 2000 – 1500 = 500 mm of Hg.
- The decrease in pressure of oxygen after 74 min $=500\times \dfrac{74}{47}=787.2\text{ }mm\text{ of }Hg$
- The remaining pressure of oxygen after 74 min = 2000 – 787.2 = 1212.8 mm of Hg.
- The formula involved in Graham’s law of diffusion is
\[\begin{align}
  & \dfrac{{{r}_{gas}}}{{{r}_{{{O}_{2}}}}}=\sqrt{\dfrac{{{M}_{{{O}_{2}}}}}{{{M}_{gas}}}} \\
 & \dfrac{{{r}_{gas}}}{{{r}_{{{O}_{2}}}}}=\dfrac{{{P}_{g}}}{{{P}_{{{O}_{2}}}}} \\
 & \sqrt{\dfrac{32}{79}}=\dfrac{{{P}_{g}}}{787.2} \\
 & {{P}_{g}}=\sqrt{\dfrac{32}{79}}\times 787.2 \\
 & {{P}_{g}}=500.8mm\text{ }Hg \\
\end{align}\]
- So, after 74 min the pressure of the other gas = 2000 – 500.8 = 1499.2 mm of Hg.
- Now we have to calculate the molar ratio of the molar ratio of the two gases remaining in the bulb after a period of 74 min.
\[\begin{align}
  & \text{molar ratio = }\dfrac{\text{pressure of the anothergas}}{\text{pressure of the oxygen}} \\
 & =\dfrac{1499.2}{1212.8} \\
 & =1.236 \\
\end{align}\]
- The molar ratio of the two gases remaining in the bulb after a period of 74 min is 1.236.

So, the correct option is B.

Note:
First we have to calculate the pressure of the each gas after 74 min by using Graham’s formula later we have to substitute the pressure of the each gas after 74 min in the molar ratio formula to get the molar ratio of the two gases remaining in the bulb after a period of 74 min.