JEE Main · 2019 · Shift-IImediumMOLE-156

8\,g of NaOH is dissolved in 18\,g of H2O. Mole fraction of NaOH in solution and molality (in mol/kg) of the solution…

Some Basic Concepts (Mole Concept) · Class 11 · JEE Main Previous Year Question

Question

8 g8\,\text{g} of NaOH is dissolved in 18 g18\,\text{g} of \ceH2O\ce{H2O}. Mole fraction of NaOH in solution and molality (in mol/kg) of the solution respectively are:

Options
  1. a

    0.167,  11.110.167,\;11.11

    ✓
  2. b

    0.167,  22.200.167,\;22.20

  3. c

    0.2,  11.110.2,\;11.11

  4. d

    0.2,  22.200.2,\;22.20

Correct Answera

0.167,  11.110.167,\;11.11

Detailed Solution

🧠 Just count moles directly 8 g8\,\text{g} of NaOH is 0.2 mol0.2\,\text{mol}. 18 g18\,\text{g} of water is 1.0 mol1.0\,\text{mol}. From these two numbers, both mole fraction and molality follow.

🗺️ Two steps Mole fraction: x\ceNaOH=0.20.2+1.0=0.21.2≈0.167.x_{\ce{NaOH}} = \frac{0.2}{0.2 + 1.0} = \frac{0.2}{1.2} \approx 0.167. Molality: m=0.20.018 kg≈11.11 m.m = \frac{0.2}{0.018\,\text{kg}} \approx 11.11\,\text{m}.

So the pair is (0.167, 11.11)(0.167,\ 11.11), matching option (a).

Answer: (a)\boxed{\text{Answer: (a)}}

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8\,g of NaOH is dissolved in 18\,g of H2O. Mole fraction of NaOH in solution and molality (in… (JEE Main 2019) | Canvas Classes