A 500 mL flask was charged with 1.0 mol of `COCl_(2)(g)` and heated to some temperature (T) where decomposes partially as , `COCl_(2)(g) hArr CO(g) + Cl_(2)(g), K_(c)(T) = 1.5 M`
Now the above flask is connected to another flask containing some pure chlorine gas at the same temperature and pressure, by a narrow tube of negligible volume. When the equilibrium was restored, the concentration `COCl_(2)(g)` was found to be 0.694 M. Determine the volume of `Cl_(2)(g)` flask.

Options

A. 200 mL
B. 100 mL
C. 150 mL
D. 120 mL

Answer

A

Explanation

`underset(1-x)(COCl_(2)(g)) hArr underset(x)(CO(g)) + underset(x)(Cl_(2)(g)), K_(c) = (2x^(2))/(1-x) = 1.5 rArr x^(2) + 0.75 x - 0.75 = 0`
`rArr x = (-0.75 + sqrt((0.75)^(2) + 3))/(2) = 0.568` mol
Now, let the volume of flask containing `Cl_(2) (g)` = V mL and it contains y mole of gas.
`rArr (V)/(y) = (500)/(1.568) = 318.87` .....(i)
`{:(COCl_(2),hArr,CO,+,Cl_(2),),(0.432 + Z,,0.568 - Z,,0.568 + y-z,):}`
`rArr [COCl_(2)] = (0.432 + Z)/(500 + V) xx 1000 = 0.694 rArr 432 + 1000 Z = 347 + 0.694 V`
`rArr Z = (0.694 V- 85)/(1000)` .....(ii)
Also : `K_(c )= 1.5 = ((0.568 + y-z)(0.568-z))/((0.432 + z)) (1000)/(500+V)` ......(iii)
Solving eqs. (i), (ii) and (iii) gives : V = 200 mL