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Some Basic Concepts of Chemistry question

2014 · Shift 1 · Q3
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Some Basic Concepts of Chemistry question

2014 · Shift 1 · Q3

JEE AdvancedChemistrySome Basic Concepts of ChemistryNumerical+3 / −1
A compound H2XH_2XH2​X with molar weight of 80g is dissolved in a solvent having density of 0.4 gml–1 . Assuming no change in volume upon dissolution, the molality of a 3.2 molar solution is
Numerical answer
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Correct answer: 8

  1. Given data

    • Solute: H2XH_2XH2​X
    • Molar mass of solute =80 g mol−1= 80\,\text{g mol}^{-1}=80g mol−1
    • Solution molarity =3.2 M= 3.2\,M=3.2M
    • Density of solvent =0.4 g mL−1= 0.4\,\text{g mL}^{-1}=0.4g mL−1
    • No change in volume on dissolution

  2. Take 1 L of solution Since molarity is 3.2 M3.2\,M3.2M, in 1 L1\,\text{L}1L of solution: moles of solute=3.2 mol\text{moles of solute} = 3.2\,\text{mol}moles of solute=3.2mol

  3. Mass of solute in 1 L solution mass of solute=3.2×80=256 g\text{mass of solute} = 3.2 \times 80 = 256\,\text{g}mass of solute=3.2×80=256g

  4. Volume of solvent used Because there is no change in volume on dissolution, the volume of solution equals the volume of solvent. So, for 1 L1\,\text{L}1L solution, solvent volume =1 L=1000 mL= 1\,\text{L} = 1000\,\text{mL}=1L=1000mL.

  5. Mass of solvent Using density: mass of solvent=0.4 g mL−1×1000 mL=400 g\text{mass of solvent} = 0.4\,\text{g mL}^{-1} \times 1000\,\text{mL} = 400\,\text{g}mass of solvent=0.4g mL−1×1000mL=400g =0.4 kg= 0.4\,\text{kg}=0.4kg

  6. Calculate molality Molality is moles of solute per kg of solvent: m=3.20.4=8 mol kg−1m = \frac{3.2}{0.4} = 8\,\text{mol kg}^{-1}m=0.43.2​=8mol kg−1

  7. Final answer 8\boxed{8}8​

  8. Comparison with stored correct answer Stored correct answer = 888

    My derived answer matches the stored answer.

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