Rakshita designed an experiment taking two clean test tubes, A and B and filled them with lime water. In test tube A, the surrounding air that we inhale was passed on by sucking air from the pipe, and in test tube B, the exhaled air was blown through the pipe. What do you think she is trying to investigate? How can she confirm her findings?
Rakshita is trying to investigate the difference in carbon dioxide content between inhaled air (surrounding air) and exhaled air.
In test tube A, the lime water will remain clear or show very slight change because inhaled air contains less carbon dioxide.
In test tube B, the lime water will turn milky or cloudy because exhaled air contains more carbon dioxide.
She can confirm her findings by observing that lime water turns milky only in the presence of carbon dioxide, and the degree of cloudiness indicates the amount of carbon dioxide present. The milky appearance in test tube B confirms that exhaled air contains more carbon dioxide than inhaled air.
Explanation
The chapter describes Activity 9.3 which demonstrates that 'lime water turns milky when it reacts with carbon dioxide.' The experiment shows that exhaled air contains more carbon dioxide than inhaled air. This is because during respiration, oxygen is used to break down glucose, and carbon dioxide is produced as a waste product. The chapter states: 'This indicates that the exhaled air contains more carbon dioxide than the air we inhale.'
Solution Steps
Rakshita is investigating the difference in carbon dioxide content between inhaled air and exhaled air.
In test tube A (inhaled air), lime water remains clear due to less carbon dioxide.
In test tube B (exhaled air), lime water turns milky due to more carbon dioxide.
Lime water turns milky only in the presence of carbon dioxide.
The milky appearance in test tube B confirms exhaled air contains more carbon dioxide than inhaled air.