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Solutions question

2024 · 8 Apr · Shift 1 · Q26
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Solutions question

2024 · 8 Apr · Shift 1 · Q26

JEE MainChemistrySolutionsNumerical+4 / −1
A solution containing 10 g10 \mathrm{~g}10 g of an electrolyte AB2\mathrm{AB}_2AB2​ in 100 g100 \mathrm{~g}100 g of water boils at 100.52∘C100.52^{\circ} \mathrm{C}100.52∘C. The degree of ionization of the electrolyte (α)(\alpha)(α) is ‾\underline{\hspace{2cm}}​×10−1\times 10^{-1}×10−1. (nearest integer) [Given : Molar mass of AB2=200 g mol−1, Kb\mathrm{AB}_2=200 \mathrm{~g} \mathrm{~mol}^{-1}, \mathrm{~K}_{\mathrm{b}}AB2​=200 g mol−1, Kb​(molal boiling point elevation const. of water) =0.52 K kg mol−1=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}=0.52 K kg mol−1, boiling point of water =100∘C;AB2=100^{\circ} \mathrm{C} ; \mathrm{AB}_2=100∘C;AB2​ ionises as AB2→A2++2 B−]\mathrm{AB}_2 \rightarrow \mathrm{A}^{2+}+2 \mathrm{~B}^{-}]AB2​→A2++2 B−]
Numerical answer
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Correct answer: 5

  1. Use boiling point elevation formula

    ΔTb=iKbm\Delta T_b = i K_b mΔTb​=iKb​m

    Given boiling point of solution =100.52∘C=100.52^\circ C=100.52∘C and pure water =100∘C=100^\circ C=100∘C,

    ΔTb=100.52−100=0.52 K\Delta T_b = 100.52-100 = 0.52\,KΔTb​=100.52−100=0.52K

  2. Calculate molality of the solution

    Mass of solute AB2=10 g\mathrm{AB_2}=10\,gAB2​=10g

    Molar mass of AB2=200 g mol−1\mathrm{AB_2}=200\,g\,mol^{-1}AB2​=200gmol−1

    moles of AB2=10200=0.05 mol\text{moles of } \mathrm{AB_2} = \frac{10}{200} = 0.05\,molmoles of AB2​=20010​=0.05mol

    Mass of water =100 g=0.1 kg=100\,g=0.1\,kg=100g=0.1kg

    m=0.050.1=0.5 mol kg−1m = \frac{0.05}{0.1} = 0.5\,mol\,kg^{-1}m=0.10.05​=0.5molkg−1

  3. Find van't Hoff factor iii

    0.52=i×0.52×0.50.52 = i \times 0.52 \times 0.50.52=i×0.52×0.5

    i=0.520.26=2i = \frac{0.52}{0.26} = 2i=0.260.52​=2

  4. Relate iii with degree of ionization α\alphaα

    AB2→A2++2B−\mathrm{AB_2 \rightarrow A^{2+} + 2B^-}AB2​→A2++2B−

    One formula unit gives total 333 ions.

    For an electrolyte giving ν=3\nu=3ν=3 ions,

    i=1+(ν−1)α=1+2αi = 1 + (\nu-1)\alpha = 1 + 2\alphai=1+(ν−1)α=1+2α

    Since i=2i=2i=2,

    2=1+2α2 = 1 + 2\alpha2=1+2α

    2α=12\alpha = 12α=1

    α=0.5=5×10−1\alpha = 0.5 = 5 \times 10^{-1}α=0.5=5×10−1

  5. Nearest integer in the blank

    5\boxed{5}5​

The derived answer matches the stored correct answer.

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