Question

3.
A vessel contains O2 and H2 in 28 : 13 molar ratio
at 100 atm pressure. Then calculate ratio of their
rate of diffusion
1. 1 :2
2. 13 :28
3. 7: 13
4. 13 :7​

Answer 1

Answer:7/13

Explanation:

Answer 2

Answer:

The ratio of the of diffusion of O2 and H2 = 7:13

Explanation:

Rate of Diffusion

• According to Graham's law, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.
• But the rate of diffusion of a gas is directly proportional to pressure.

That is, the rate of diffusion of a gas (r) = Pressure (P) /$\sqrt{Molar Mass (M)}$

From the ideal gas equation, the pressure (P) ∝ number of moles of gas (n)

∴ The rate of diffusion (r) = n/$\sqrt{M}$

r$O_{2}$ / r$H_{2}$  = (n$O_{2}$ / n$H_{2}$) ÷ $\sqrt{M_{O_{2} } }$ /$\sqrt{M_{H_{2} } }$)

               = (n$O_{2}$ / n$H_{2}$) x $\sqrt{M_{H_{2} } }$ /$\sqrt{M_{O_{2} } }$)

               

Given that, (n$O_{2}$ / n$H_{2}$) = 28:13

∴ r$O_{2}$ / r$H_{2}$ = $\frac{28}{13}$ x  $\sqrt{\frac{2}{32} }$

                 =  $\frac{28}{13}$ x $\frac{1}{4}$

                 = $\frac{7}{13}$

∴ The ratio of the of diffusion of O2 and H2 (r$O_{2}$ / r$H_{2}$) = 7:13