Five millitres of a gas (A) containing only C and H was mixed with an excess of oxygen (30 ml) and the mixture was exploded by means of an electric spark. After the explosion, the remaining volume of the mixed gases was 25 ml. On adding a concentrated solution of KOH, the volume further diminished to 15 ml.The residual gas being pure oxygen . The molecule formula of gas (A) isA. `C_2H_4`B. `C_2H_6`C. `C_3H_6`D. `C_3H_8`

Five millitres of a gas (A) containing only C and H was mixed with an

1.	Five millitres of a gas (A) containing only C and H was mixed with an excess of oxygen (30 ml) and the mixture was exploded by means of an electric spark. After the explosion, the remaining volume of the mixed gases was 25 ml. On adding a concentrated solution of KOH, the volume further diminished to 15 ml.The residual gas being pure oxygen . The molecule formula of gas (A) isA. `C_2H_4`B. `C_2H_6`C. `C_3H_6`D. `C_3H_8`
Answer» Correct Answer - A `C_x H_y(g)+(x+y/4)O_2(g) to xCO_2(g)+y/2H_2O(1)` 1 ml `(x+y/4)` ml x ml- 5 ml `5(x+y/4)` ml 5x ml - Volume of gases remaining (i.e. volume of `CO_2` + volume of `O_2` unreacted)=25 ml Volume remaining after absorbing KOH=15 ml Volume absorbed by KOH=Volume of `CO_2`=25-15=10 ml Volume of `CO_2 rArr ` 5x =10 ml , x=2 Volume of `O_2` taken =30 ml Volume of `O_2` left =15 ml Volume of `O_2` reached =30-15=15 ml `therefore 5(x+y/4)=15 rArr 5(2+y/4)=15 rArr y=4` Formula of gas (A)=`C_2H_4` (Ethene)