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Some Basic Concepts of Chemistry question

2019 · 9 Apr · Shift 2 · Q19
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Some Basic Concepts of Chemistry question

2019 · 9 Apr · Shift 2 · Q19

JEE MainChemistrySome Basic Concepts of ChemistryMCQ+4 / −1
What would be the molality of 20% (mass/ mass) aqueous solution of KI? (molar mass of KI = 166 g mol–1)
  1. A
    1.51
  2. B
    1.35
  3. C
    1.08
  4. D
    1.48
View written solutionFree

Correct answer: A

  1. Interpret 20% (mass/mass) aqueous solution

A 20%20\%20% (w/w)(w/w)(w/w) solution means:

  • 20 20\,20g of KI is present in
  • 100 100\,100g of solution

So, mass of water (solvent) is: 100−20=80 g=0.080 kg100 - 20 = 80\,\text{g} = 0.080\,\text{kg}100−20=80g=0.080kg

  1. Calculate moles of KI

Given molar mass of KI = 166 g mol−1166\,\text{g mol}^{-1}166g mol−1

moles of KI=20166=0.1205 mol\text{moles of KI} = \frac{20}{166} = 0.1205\,\text{mol}moles of KI=16620​=0.1205mol

  1. Use formula for molality

Molality is defined as: m=moles of solutemass of solvent in kgm = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}m=mass of solvent in kgmoles of solute​

Thus, m=0.12050.080=1.50625m = \frac{0.1205}{0.080} = 1.50625m=0.0800.1205​=1.50625

  1. Round off

m≈1.51m \approx 1.51m≈1.51

  1. Check options
  • A: 1.511.511.51 ✔️
  • B: 1.351.351.35
  • C: 1.081.081.08
  • D: 1.481.481.48

So, the correct option is A.

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