# CBSE Class 11-science Answered

**10 ml of mixture of CH4, C2H4 and CO2 are exploded with excess O2. A contraction of 17ml was observed after explosion. After the treatment with KOH solution there was a 2nd contraction of 14ml. Find the % of gases in mixture**

Let the volume of CH_{4} = x ml and volume of C_{2}H_{4} = y ml

Hence volume of CO_{2} = (10 - x - y) ml

Since after the explosion, there was contraction of 17ml; this means that when the explosion takes place at that time water is in gaseous state, after the explosion water vapours converted into liquid state and the volume of liquid water is assumed to be zero.

Hence 2x + 2y = 17 .. (a)

Since KOH absorbs CO_{2} gas when the gaseous mixture is treated with KOH further reduction in volume is 14 ml, hence volume of CO_{2} in gaseous mixture is 14 ml.

10 - x - y + x + 2y = 14 .. (b)

10 + y = 14

Y = 4 ml

On putting this value in equation (a), we get

x = 4.5

So, volume of CH_{4} = 4.5 ml

Volume of C_{2}H_{4} = 4 ml

Volume of CO_{2} = (10 - 4 - 4.5) = 1.5 ml