Match List - I with List – II and select the correct option (Based on mole concept):
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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".
Step 1: Analyze each item in List-I and find its corresponding value in List-II. (a) 2 moles of ethene (C\(_2\)H\(_4\)):
Molar mass of ethene = (2 \(\times\) 12.01) + (4 \(\times\) 1.01) \(\approx\) 28 g/mol.
Mass of 2 moles of ethene = 2 mol \(\times\) 28 g/mol = 56 g.
This matches with (ii) 56 g. So, a \(\rightarrow\) ii. (b) Molar mass is equal to 66 g:
This statement is poorly phrased. It likely means "The mass of the sample is 66 g". Let's check List-II for an option with a mass of 66 g.
Consider (iv) 1.5 mole of CO\(_2\):
Molar mass of CO\(_2\) = 12 + (2 \(\times\) 16) = 44 g/mol.
Mass of 1.5 moles of CO\(_2\) = 1.5 mol \(\times\) 44 g/mol = 66 g.
This matches. So, b \(\rightarrow\) iv. (c) 1 g of H\(_2\):
Molar mass of H\(_2\) = 2 g/mol.
Moles of H\(_2\) = mass / molar mass = 1 g / 2 g/mol = 0.5 moles.
Volume of 0.5 moles of any gas at STP = 0.5 mol \(\times\) 22.4 L/mol = 11.2 L.
This matches with (i) 11.2 L volume at STP. So, c \(\rightarrow\) i. (d) 2 moles of water vapours (H\(_2\)O):
Number of molecules in 2 moles = 2 mol \(\times\) Avogadro's number (N\(_A\))
= 2 \(\times\) 6.022 \(\times\) 10\(^{23}\) molecules = 12.044 \(\times\) 10\(^{23}\) molecules.
This matches with (iii) 12.04 \(\times\) 10\(^{23}\) molecules. So, d \(\rightarrow\) iii. Step 2: Compile the matches.
The correct matches are: a \(\rightarrow\) ii, b \(\rightarrow\) iv, c \(\rightarrow\) i, d \(\rightarrow\) iii. Step 3: Final Answer:
This combination corresponds to option (A).
