A mixture of 100 ml of CO, CO2 and O2 was sparked.When the resulting gaseous mixture was passed through KOH solution,contraction of volume was found to be 80 ml. The composition of initial mixture may be(in the same order):

a) 30 ml, 60 ml, 10 ml

b) 30 ml, 50 ml, 20 ml

c) 50 ml, 30 ml, 20 ml

d) 20 ml, 70 ml, 20 ml

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Let, initially there was x ml CO, and y ml O2, then volume of CO2 was initially [100-(x+y)] ml

Upon sparking, 2 vol. CO reacts with 1 vol. O2 to produce 2 vol. CO2

So, x vol. CO reacts with x/2 vol. O2 to produce x vol. CO2

After sparking, total CO2 is thus [100-(x+y)] + x = 100-y ml

Upon passing through KOH solution, the entire amount of CO2 (g) from a gas mixture gets absorbed, which is exhibited as the contraction in volume there on.

Here, contraction is found to be 80 ml

So, 100-y = 80

Or, y = volume of O2 = 100-80 = 20 ml.

So, the option a) is discarded.

Option d) makes total gas mixture more than 100 ml, hence discarded.

Option c) has 50 ml CO which upon oxidation can produce 50 ml CO2 totalling 80 ml CO2 upon sparking, but it requires 50/2 = 25 ml oxygen which is not present here, Therefore the correct answer is option b) 30 ml, 50 ml, 20 ml of CO, CO2 and O2 respectively.

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### 1 Reply | Hide Replies ▲

Respected teacher,

The answers in the book are both b) and c).

What to do now?

I had suggested to discard option c) stating reasons for doing so in my answer.

Furthermore, how can somebody get 80 ml CO2 from the combination stated in option c)? It can produce at best 30 ml(originally)+40 ml(obtained from sparking with 20 ml O2 and sufficient CO) = 70 ml CO2.

So, there must be some thing wrong in the answer which suggests both b) and c).

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