`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
Text Solution
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The correct Answer is:
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`
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