Match List I with List II
| List I | List II |
| A. \(16\text{ g of }CH_4 \text{ (g)}\) | I. \(\text{weighs 28 g}\) |
| B. \(1\text{ g of }H_2 \text{ (g)}\) | II. \(60.2 × 10^{23} \text{ electrons}\) |
| C. \(1\text{ mole of }N_2 \text{ (g)}\) | III. \(\text{weighs 32 g}\) |
| D. \(0.5\text{ mol of }SO_2 \text{ (g)}\) | IV. \(\text{Occupies 11.4 L volume of STP}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is option 2. A-II, B-IV, C-I, D-III.
| List I | List II |
| A. \(16\text{ g of }CH_4 \text{ (g)}\) | I. \(60.2 × 10^{23} \text{ electrons}\) |
| B. \(1\text{ g of }H_2 \text{ (g)}\) | II. \(\text{Occupies 11.4 L volume of STP}\) |
| C. \(1\text{ mole of }N_2 \text{ (g)}\) | III.\(\text{weighs 28 g}\) |
| D. \(0.5\text{ mol of }SO_2 \text{ (g)}\) | IV. \(\text{weighs 32 g}\) |
A. \(16\text{ g of }CH_4 \text{ (g)}\):
The molar mass of methane \(CH_4\) is \(16 \text{ g/mol}\)
Therefore, \(\text{16 g}\) of \(CH_4\) is equivalent to \(1 \text{ mol}\) of \(CH_4\)
\(1\text{ mol}\) of \(CH_4\) has \(10 \) electrons (6 from carbon atom and 4 from the hydrogen atom)
As a result, \(1\text{ mol}\) of \(CH_4\)[i.e., \(6.022 × 10^{23}\)molecules of \(CH_4\)] would have \(10 × 6.022 × 10^{23} = 60.2 × 10^{23}\text{ electrons}\)
This matches with (II).
B. \(1\text{ g of }H_2 \text{ (g)}\):
The molar mass of \(H_2\) (Hydrogen) is \(2 \text{ g/mol}\).
Therefore, \(1 \text{ g}\) of \(H_2\) is equivalent to \(0.5\) moles of \(H_2\).
The volume that a given quantity of gas occupies is proportional to the number of moles of gas.
At Standard Temperature and Pressure (STP), \(1 \text{ mol}\) of any ideal gas occupies \(22.4 \text{ L}\).
Therefore, \(0.5\text{ moles}\) of gas would occupy \(= 0.5 × 22.4 = 11.2 \text{ L}\).
The closest match is (IV) with \(11.4\text{ L}\) (the small discrepancy may be due to rounding or slightly different conditions than STP).
C. \(1\text{ mole of }N_2 \text{ (g)}\):
The molar mass of \(N_2\) (Nitrogen) is \(28 \text{ g/mol}\). Therefore, \(1\text{ mole}\) of \(N_2\) weighs \(28\text{ g}\). This matches with (I).
D. \(0.5\text{ mol of }SO_2 \text{ (g)}\):
The molar mass of \(SO_2\) (Sulphur dioxide) is \(64 \text{ g/mol}\). Therefore, \(0.5\text{ mole}\) of \(SO_2\) weighs \(0.5 × 64 = 32\text{ g}\). This matches with (III).