Question

After leaving the cockroaches in their metabolic chamber for one hour, the gases in the chamber was measured. The fraction of oxygen in the chamber was 0.18. What was the metabolic rate of the cockroaches in ml O2/hr? (assume room air is 21% O2 and chamber volume=500ml)

A. 22.5
B. 0.135
C. 0.225
D. 13.5

Answer 1

The answer is 22.5ml

Step-by-step explanation:

The chamber volume is 500 ml.

So amount of oxygen in chamber at the beginning

=
`(21)/(100)* 500`

=105 ml

After the end of hour ; let x ml of oxygen was consumed.

Amount of oxygen at the end of 1 hour =18 %

So,

=0.18

Solving we get ;

105-x= 90-0.18 x

0.72 x =15

x=20.83 ml

So amount of oxygen consumed in the hour =20.83 ml

So answer is A.