Correct Answer - A
As partial pressure of a gas in a mixture is the pressure it would exert for the same volume and temperature, if it alone occupied the vessel, therefore, for common V and T, `P_(1) and P_(2)` are partial pressures.
`P_(1)V=mu_(1)RT andP_(2)V=mu_(2)RT.`
Here, 1 and 2 refer to neon gas and oxygen gas respectively.
Now, `(P_(1))/(P_(2))=(mu_(1))/(mu_(2))implies(mu_(1))/(mu_(2))=3/2("Given"(P_(1))/(P_(2))=(3)/(2))`
If `N_(1) and N_(2)` are number of molecules of two gases, then
`therefore(N_(1))/(N_(2))=(mu_(1))/(mu_(2))=3/2,Where mu_(1)=(N_(1))/(N_(A))and mu_(2)=(N_(2))/(N_(A))`