Question

Arrange the following in increasing order of the volume (in L) occupied by them at STP (i) 1.5 moles of \(CO_2\)
(ii) 14 g of \(N_2\)
(iii) \(10^{21}\) molecules of oxygen

• A. (iii) < (ii) < (i)
• B. (i) < (ii) < (iii)
• C. (ii) < (i) < (iii)
• D. (i) < (iii) < (ii)

Hint:

At STP, just compare moles — volume directly follows.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
At Standard Temperature and Pressure (STP), 1 mole of any ideal gas occupies a volume of 22.4 L. To compare the volumes, we should convert all given quantities into moles.
Step 2: Key Formula or Approach:
1. Moles from mass: \( n = \frac{\text{mass}}{\text{molar mass}} \).
2. Moles from molecules: \( n = \frac{N}{N_A} \).
3. Volume at STP: \( V = n \times 22.4 \text{ L} \).
Step 3: Detailed Explanation:
(i) 1.5 moles of \(CO_2\):
Volume \( V_1 = 1.5 \times 22.4 = 33.6 \text{ L} \).

(ii) 14 g of \(N_2\):
Molar mass of \(N_2 = 28 \text{ g/mol} \).
Moles \( n = 14 / 28 = 0.5 \text{ mol} \).
Volume \( V_2 = 0.5 \times 22.4 = 11.2 \text{ L} \).

(iii) \(10^{21}\) molecules of oxygen:
Avogadro's number \( N_A \approx 6.02 \times 10^{23} \).
Moles \( n = \frac{10^{21}}{6.02 \times 10^{23}} \approx 0.00166 \text{ mol} \).
Volume \( V_3 = 0.00166 \times 22.4 \approx 0.037 \text{ L} \).

Comparing the values: \( V_3 (0.037)<V_2 (11.2)<V_1 (33.6) \).
Step 4: Final Answer:
The increasing order is (iii) < (ii) < (i).