In the following diagram,container of `NH_(3)` gas and container of HCl gas, connected through a long tube, are opened simultaneously at both ends, the white `NH_(4) Cl` ring first formed will be at Q point. If `OP = 40 cm`, then find OQ -

Options

A. 35 cm
B. 23.74 cm
C. 30 cm
D. 31.25 cm

Answer

B

Explanation

Let OQ =x cm so QP = ( 40 - x ) cm
Diffused volume of `NH_(3)` gas =Area of T.S. of tube `xx` Distance travelled by `NH_(3)` gas
`V_(NH_(3)) =A xx OQ = Ax`
{Where A is area of T.S. of tube }
Similarly in the same time,
Diffused volume of HCl gas = Area of T.S. of tube `xx` Distance travelled by HCl gas
`V_(HCl ) = A xx QP = A ( 40 - x ) `
From Graham's Law of diffusion
`rArr ( r_(NH_(3)))/(r_(HCl))= sqrt((M_(HCl))/(M_(NH_(3)))) `
`rArr ( V_(NH_(3)//t))/(V_(HCl//t)) =sqrt((36.5)/( 17)) = 1.46`
`rArr ( x)/(( 40-x)) =1.46 rArr x = 23. 74 cm`
`:. OQ = 23.74 cm`