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Solutions question

2021 · 25 Jul · Shift 2 · Q23
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Solutions question

2021 · 25 Jul · Shift 2 · Q23

JEE MainChemistrySolutionsNumerical+4 / −1
When 3.00 g of a substance 'X' is dissolved in 100 g of CCl4CCl_4CCl4​, it raises the boiling point by 0.60 K. The molar mass of the substance 'X' is ‾\underline{\hspace{2cm}}​ g mol −-− 1. (Nearest integer). [Given Kb for CCl4CCl_4CCl4​ is 5.0 K kg mol −-− 1]
Numerical answer
View written solutionFree

Correct answer: 250

  1. Use boiling point elevation formula

    ΔTb=Kbm\Delta T_b = K_b mΔTb​=Kb​m

    where:

    • ΔTb=0.60 K\Delta T_b = 0.60\,\text{K}ΔTb​=0.60K
    • Kb=5.0 K kg mol−1K_b = 5.0\,\text{K kg mol}^{-1}Kb​=5.0K kg mol−1
    • mmm = molality

  2. Calculate molality

    m=ΔTbKb=0.605.0=0.12 mol kg−1m = \frac{\Delta T_b}{K_b} = \frac{0.60}{5.0} = 0.12\,\text{mol kg}^{-1}m=Kb​ΔTb​​=5.00.60​=0.12mol kg−1

  3. Mass of solvent in kg

    Given solvent = 100 g=0.100 kg100\,\text{g} = 0.100\,\text{kg}100g=0.100kg

  4. Calculate moles of solute

    m=moles of solutekg of solventm = \frac{\text{moles of solute}}{\text{kg of solvent}}m=kg of solventmoles of solute​

    So,

    moles of solute=m×kg of solvent=0.12×0.100=0.012 mol\text{moles of solute} = m \times \text{kg of solvent} = 0.12 \times 0.100 = 0.012\,\text{mol}moles of solute=m×kg of solvent=0.12×0.100=0.012mol

  5. Calculate molar mass of solute

    Molar mass=massmoles=3.000.012=250 g mol−1\text{Molar mass} = \frac{\text{mass}}{\text{moles}} = \frac{3.00}{0.012} = 250\,\text{g mol}^{-1}Molar mass=molesmass​=0.0123.00​=250g mol−1

  6. Nearest integer

    250250250

Final Answer

250\boxed{250}250​

The derived answer matches the stored correct answer.

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