Question

Match List - I with List – II and select the correct option (Based on mole concept):

Codes:

• A. a – ii, b – iv, c – i, d – iii
• B. a – iii, b – i, c – iv, d – ii
• C. a – i, b – iv, c – ii, d – iii
• D. a – ii, b – iii, c – i, d – iv

Hint:

In matching questions, if a statement seems ambiguous (like option b), solve the clear-cut matches first (a, c, d). This will help you deduce the correct match for the ambiguous item by elimination. Here, "Molar mass is equal to 66 g" was a confusing way to say "The mass is 66 g".

Answer 1

The Correct Option is A

To solve this question, we need to match the items in List-I with their corresponding items in List-II using concepts from the mole concept.

1. (a) 2 moles of ethene 
   This can be related to \(11.2 \, \text{L}\) because at Standard Temperature and Pressure (STP), 1 mole of any gas occupies \(22.4 \, \text{L}\). Therefore, 2 moles would occupy \(2 \times 22.4 = 44.8 \, \text{L}\), hence not matching any given volume directly.
2. (b) Molar mass is equal to 66 g 
   This would correctly correspond to \(56 \, \text{g}\) in CO₂ because 1 mole of CO₂ is approximately 44 g/mol, equating closer elements or similar compounds configurations.
3. (c) 1 g of H₂ 
   1 gram of H₂ contains \(\frac{1}{2} \, \text{mole of H}_2\). The number of molecules is \(0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23}\), which doesn't match any in List-II directly.
4. (d) 2 moles of water vapours 
   This corresponds to \(1.5 \, \text{mole of CO}_2\) molecules strongly forming 1.5 relationships as comparisons.

Let's match with the correct List-II items:

• (a) 2 moles of ethene — 12.04 × 10²³ molecules (ii)
• (b) Molar mass is equal to 66 g — 1.5 moles of CO₂ (iv)
• (c) 1 g of H₂ — 11.2 L volume at STP (i)
• (d) 2 moles of water vapours — 56 g (iii)

The correct option is: a – ii, b – iv, c – i, d – iii.