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Some Basic Concepts of Chemistry question

2023 · 10 Apr · Shift 2 · Q11
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Some Basic Concepts of Chemistry question

2023 · 10 Apr · Shift 2 · Q11

JEE MainChemistrySome Basic Concepts of ChemistryMCQ+4 / −1

Match List I with List II

List - I List - II (Δ0\Delta_0Δ0​)
A. 16 g of CH4 (g)\mathrm{CH_4~(g)}CH4​ (g) I. Weighs 28 g
B. 1 g of H2 (g)\mathrm{H_2~(g)}H2​ (g) II. 60.2×102360.2\times10^{23}60.2×1023 electrons
C. 1 mole of N2 (g)\mathrm{N_2~(g)}N2​ (g) III. Weighs 32 g
D. 0.5 mol of SO2 (g)\mathrm{SO_2~(g)}SO2​ (g) IV. Occupies 11.4 L volume of STP

Choose the correct answer from the options given below:

  1. A
    A-II, B-III, C-IV, D-I
  2. B
    A-II, B-IV, C-I, D-III
  3. C
    A-I, B-III, C-II, D-IV
  4. D
    A-II, B-IV, C-III, D-I
View written solutionFree

Correct answer: B

  1. Find the properties of each item in List I

We use:

  • 111 mole gas at STP occupies 22.4 L22.4\,\text{L}22.4L
  • Avogadro number NA=6.022×1023N_A = 6.022\times 10^{23}NA​=6.022×1023
  1. Option A: 16 g of CH4(g)16\,\text{g of } \mathrm{CH_4(g)}16g of CH4​(g)

Molar mass of CH4\mathrm{CH_4}CH4​: 12+4(1)=16 g mol−112 + 4(1) = 16\,\text{g mol}^{-1}12+4(1)=16g mol−1

So, 16 g16\,\text{g}16g of CH4\mathrm{CH_4}CH4​ means 111 mole of CH4\mathrm{CH_4}CH4​.

Each molecule of CH4\mathrm{CH_4}CH4​ has:

  • Carbon: 666 electrons
  • Hydrogen: 4×1=44 \times 1 = 44×1=4 electrons

Total electrons per molecule =10= 10=10.

Thus, 111 mole of CH4\mathrm{CH_4}CH4​ contains: 10NA=10(6.022×1023)=60.22×1023 electrons10N_A = 10(6.022\times 10^{23}) = 60.22\times 10^{23} \text{ electrons}10NA​=10(6.022×1023)=60.22×1023 electrons

So, A→IIA \to IIA→II

  1. Option B: 1 g of H2(g)1\,\text{g of } \mathrm{H_2(g)}1g of H2​(g)

Molar mass of H2\mathrm{H_2}H2​: 2 g mol−12\,\text{g mol}^{-1}2g mol−1

Therefore, 1 g1\,\text{g}1g of H2\mathrm{H_2}H2​ is: 12 mole\frac{1}{2} \text{ mole}21​ mole

At STP, volume occupied: 0.5×22.4=11.2 L0.5 \times 22.4 = 11.2\,\text{L}0.5×22.4=11.2L

The table says 11.4 L11.4\,\text{L}11.4L, which is evidently intended to represent half the molar volume at STP (likely a rounding/printing issue; in many exam keys this pairing is taken as the STP-volume match for 0.50.50.5 mol).

Hence, B→IVB \to IVB→IV

  1. Option C: 111 mole of N2(g)\mathrm{N_2(g)}N2​(g)

Molar mass of N2\mathrm{N_2}N2​: 14×2=28 g14\times 2 = 28\,\text{g}14×2=28g

So, C→IC \to IC→I

  1. Option D: 0.50.50.5 mol of SO2(g)\mathrm{SO_2(g)}SO2​(g)

Molar mass of SO2\mathrm{SO_2}SO2​: 32+2(16)=64 g mol−132 + 2(16) = 64\,\text{g mol}^{-1}32+2(16)=64g mol−1

Mass of 0.50.50.5 mol: 0.5×64=32 g0.5 \times 64 = 32\,\text{g}0.5×64=32g

So, D→IIID \to IIID→III

  1. Final matching

Thus the correct match is: A−II,  B−IV,  C−I,  D−IIIA-II,\; B-IV,\; C-I,\; D-IIIA−II,B−IV,C−I,D−III

This corresponds to Option B.

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