20 ml of a mixture of ${CO}_{2}$ and $C_{2} H_{4}$ was mixed with excess of $O_{2}$ gas and was exploded. On bringing the solution back to the original room temperature a contraction of 12 ml was observed. What is the volume percentage of ${CO}_{2}$ in the original mixture?

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$30 %$

$\begin{aligned} 14 % \end{aligned}$

$70 %$

$6 %$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

In the reaction we have,

$\begin{aligned} \underset{x}{C_{2} H_{4}} + \underset{3 x}{3 O_{2}} ⟶ 2 {CO}_{2} (g) + 2 H_{2} O (l) \\ 2 x 2 x = 12 \\ x = 6 \end{aligned}$

$% {CO}_{2} = \frac{14}{20} \times 100 = 70 %$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE