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Osmolarity of 0.02 M potassium ferrocyanide solution at 300 K is (assume solute is 100% ionised)
  • a)
    0.10 osmol
  • b)
    0.20 osmol
  • c)
    0.02 osmol 
  • d)
    0.004 osmol 
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Osmolarity of 0.02 M potassium ferrocyanide solution at 300 K is (assu...
 Osmotic pressure is a colligative property, hence osmolarity is the molarity of the solution involving van't Hoff factor
Thus, osmolarity = 0.02 x i= 0.02 x 5 = 0.I0 osmol
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Most Upvoted Answer
Osmolarity of 0.02 M potassium ferrocyanide solution at 300 K is (assu...
Osmolarity is a measure of the concentration of solute particles in a solution. It is expressed in osmoles per liter (osmol/L). To calculate the osmolarity of a solution, we need to know the concentration of the solute and the number of particles it dissociates into.

In this case, we are given a 0.02 M potassium ferrocyanide solution. Potassium ferrocyanide, K4Fe(CN)6, dissociates into four potassium ions (K+) and one ferrocyanide ion ([Fe(CN)6]4-).

To calculate the osmolarity, we need to calculate the total number of particles in the solution. Since the solute is 100% ionized, we can assume that each molecule dissociates completely into its constituent ions.

The concentration of the potassium ions is 4 times the concentration of the potassium ferrocyanide solute, i.e., 4 * 0.02 M = 0.08 M. The concentration of the ferrocyanide ion is the same as the concentration of the potassium ferrocyanide solute, i.e., 0.02 M.

Now, we can calculate the osmolarity using the formula:

Osmolarity = (concentration of potassium ions * number of potassium ions) + (concentration of ferrocyanide ion * number of ferrocyanide ions)

Plugging in the values, we get:

Osmolarity = (0.08 M * 4) + (0.02 M * 1)
= 0.32 osmol + 0.02 osmol
= 0.34 osmol

Therefore, the osmolarity of the 0.02 M potassium ferrocyanide solution is 0.34 osmol. Since the answer choices do not include this value, we can conclude that there might be a mistake in the answer options provided.

Please note that the correct answer cannot be determined based on the information given in the question.
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Osmolarity of 0.02 M potassium ferrocyanide solution at 300 K is (assume solute is 100% ionised)a)0.10 osmolb)0.20osmolc)0.02osmold)0.004osmolCorrect answer is option 'A'. Can you explain this answer?
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Osmolarity of 0.02 M potassium ferrocyanide solution at 300 K is (assume solute is 100% ionised)a)0.10 osmolb)0.20osmolc)0.02osmold)0.004osmolCorrect answer is option 'A'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Osmolarity of 0.02 M potassium ferrocyanide solution at 300 K is (assume solute is 100% ionised)a)0.10 osmolb)0.20osmolc)0.02osmold)0.004osmolCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Osmolarity of 0.02 M potassium ferrocyanide solution at 300 K is (assume solute is 100% ionised)a)0.10 osmolb)0.20osmolc)0.02osmold)0.004osmolCorrect answer is option 'A'. Can you explain this answer?.
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