Science — CBSE Class 10 board question

(i) Exhaled air contains a much higher concentration of CO₂ (about 4%) compared to atmospheric air (about 0.04%). Lime water turns milky due to the reaction:

$$\text{Ca(OH)}_2 + \text{CO}_2 \rightarrow \text{CaCO}_3\downarrow + \text{H}_2\text{O}$$

Since exhaled air has approximately 100 times more CO₂ than atmospheric air, the lime water in diagram (II) turns milky much faster than in diagram (I). The syringe in diagram (I) pushes in air with very little CO₂, so it takes considerably longer for the lime water to turn milky.

(ii)

Diagram of Open Stomatal Pore:

```
[Guard Cell] [Guard Cell]
(I) (I)
___________ ___________
| • • • | | • • • |
| Chloroplast| |Chloroplast|
|___(II)____| PORE |___(II)____|
←— Stomatal opening —→
```

(Label: (I) Guard cells, (II) Chloroplast)

Two functions of stomata:

1. Exchange of gases (CO₂ and O₂) between the plant and the atmosphere.
2. Transpiration — loss of water vapour through stomatal pores.

Source: Life Processes, Chapter 5, Section 5.3

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• For part (i), the key idea is the difference in CO₂ concentration between atmospheric and exhaled air. Examiners expect you to state this clearly and may award marks for the chemical equation of lime water turning milky.
• For part (ii), draw a simple neat diagram showing two bean-shaped guard cells with dots representing chloroplasts inside them and a pore between them. Labels must be clear. The two functions (gas exchange and transpiration) are standard textbook points — both must be stated to get full marks.