Question

`0.01` mol of a gaseous compound `C_(2)H_(2)O_(x)` was treated with `224 mL` of `O_(2)` at `STP`. After combustion the total volume of the gases is `560 mL` at `STP`. On treatment with `KOH` solution the volume decreases to `112 mL`. The volue of `x` is:
A. `4`
B. `2`
C. `3`
D. None of these

Answer 1

Correct Answer - A
vol.of `CO_(2) +` vol. of remaining oxygen `= 560 ml`.
or vol. of remaining oxygen `= 112 mL`
`C_(2)H_(2)O_(x)(g) + ((5-x)/(2))O_(2)(g)overset(Delta)rarr2CO_(2)(g) + H_(2)O(l)`
`{:(224 mL,224 mL,0,-),(0,112 mL,448 mL,-):}`
for used mole of a`O_(2)`:
`((5-x)/(2)) xx 0.1 = (0.01)/(2)`
or `x = 4`