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120 gm of urea are present in 5 litre solution, the active mass of urea is
  • a)
    0.2
  • b)
    0.06
  • c)
    0.4
  • d)
    0.08
Correct answer is option 'C'. Can you explain this answer?
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120 gm of urea are present in 5 litre solution, the active mass of ure...
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120 gm of urea are present in 5 litre solution, the active mass of ure...
Given:
Amount of urea = 120 gm
Volume of solution = 5 L

To find: Active mass of urea

Formula used: Active mass = Amount of solute / Volume of solution

Calculation:
Active mass of urea = 120 gm / 5 L
Active mass of urea = 24 gm/L

We know that 1 mole of urea = 60 gm
Therefore, number of moles of urea = 120 gm / 60 gm = 2 moles

Active mass of urea = Number of moles of urea / Volume of solution
Active mass of urea = 2 moles / 5 L
Active mass of urea = 0.4 mol/L

Therefore, the correct option is (c) 0.4.
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120 gm of urea are present in 5 litre solution, the active mass of ure...
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120 gm of urea are present in 5 litre solution, the active mass of urea isa)0.2b)0.06c)0.4d)0.08Correct answer is option 'C'. Can you explain this answer?
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