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Some Basic Concepts of Chemistry question

2019 · 9 Jan · Shift 1 · Q12
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Some Basic Concepts of Chemistry question

2019 · 9 Jan · Shift 1 · Q12

JEE MainChemistrySome Basic Concepts of ChemistryMCQ+4 / −1
A solution of sodium sulphate contains 92 g of Na+Na^+Na+ ions per kilogram of water. The molality of Na+Na^+Na+ ions in that solution in mol kg −-− 1 is :
  1. A
    12
  2. B
    4
  3. C
    8
  4. D
    16
View written solutionFree

Correct answer: B

  1. Given data
  • Mass of Na+Na^+Na+ ions =92 g= 92\,\text{g}=92g
  • Mass of water =1 kg= 1\,\text{kg}=1kg

We need the molality of Na+Na^+Na+ ions.

  1. Formula for molality
Molality=moles of solutemass of solvent in kg\text{Molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}Molality=mass of solvent in kgmoles of solute​

Here, solute species considered is Na+Na^+Na+.

  1. Calculate moles of Na+Na^+Na+ ions

Molar mass of Na+Na^+Na+ is same as sodium atom mass:

M(Na+)=23 g mol−1M(Na^+) = 23\,\text{g mol}^{-1}M(Na+)=23g mol−1

So,

moles of Na+=9223=4\text{moles of } Na^+ = \frac{92}{23} = 4moles of Na+=2392​=4
  1. Calculate molality

Since solvent mass is 1 kg1\,\text{kg}1kg,

m=41=4 mol kg−1m = \frac{4}{1} = 4\,\text{mol kg}^{-1}m=14​=4mol kg−1
  1. Match with options
  • A: 121212
  • B: 444
  • C: 888
  • D: 161616

Hence, the correct option is:

B: 4\boxed{\text{B: }4}B: 4​
  1. Comparison with stored correct answer

Stored correct answer is B.

My derived answer is also B.

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