A 20 mL mixture of CO, `CH_4`, and Helium (He) gases is exploded by an electric discharge at room temperature with excess of oxygen. The volume contraction is found to be 13 mL. A further contraction of 14 mL occurs when the residual gas is treated wityh KOH solution. Find out the composition of the gaseous mixture in terms of volume percentage.

Options

A. `CO-40%, CH_(4)-20%, He-40%`
B. `CO-40%, CH_(4)-30%, He-20%`
C. `CO-50%, CH_(4)-20%, He-30%`
D. `CO-50%, CH_(4)-10%,He-40%`

Answer

C

Explanation

`underset(a vol)underset(1 vol)(CO(g))+((1)/(2))underset(1//2a vol)underset(1//2vol)(O_(2))(g)rarrunderset(a vol)underset(1vol)(CO_(2))(g)`
`underset(b vol)underset(1vol)(CH_(4)(g)) +underset(2b vol)underset(2vol)(2O_(2))(g)rarrunderset(b vol)underset(1 vol)(CO_(2))(g)+2H_(2)O(l)`
Total initial volume `=[a+((1)/(2))a+b+2b+20-(a+b)]`
volume after reaction `=[a+b +20 - (a+b)]`
Contraction `=((1)/(2)) a+2b = 13`
When treated with `KOH, CO_(2)` is absorbed i.e. ` a+ b = 14`
Solving both equations `a =10, b = 4`
`CO = 10 cm^(3), CH_(4) = 4cm^(3)` and `He = 6 cm^(3)` i.e., `CO = 50%, CH_(4)= 20%` and `He = 30%`