Practicing Success
Match List I with List II
Choose the correct answer from the options given below: |
A-II, B-III, C-IV, D-I A-II, B-IV, C-I, D-III A-I, B-III, C-II, D-IV A-II, B-IV, C-III, D-I |
A-II, B-IV, C-I, D-III |
The correct answer is option 2. A-II, B-IV, C-I, D-III.
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). |