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Freezing point of a biological fluid is - 0.56° C in aqueous solution.Kf (H20) = 1 .86° mol-1 kg. Thus, its osmotic pressure at 310 K is (assume molarity = molality)
- a)0.0766 atm
- b)7.66 atm
- c)0.766 atm
- d)0.914 atm
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Freezing point of a biological fluid is - 0.56° C in aqueous solut...
(b)
ΔTf = molality x Kf
0.56° = molality x 1,86° mol-1 kg
∴ Molarity = 
mol kg-1
or molarity = 0.301 mol L-1
mol kg-1or molarity = 0.301 mol L-1
Also, π = M RT= 0.301 x 0,0821 x 3 1 0 = 7.66 atm
Most Upvoted Answer
Freezing point of a biological fluid is - 0.56° C in aqueous solut...
°C, what is the fluid's osmolarity?
To answer this question, we need to use the freezing point depression equation:
ΔTf = Kf * m
Where ΔTf is the change in freezing point, Kf is the freezing point depression constant for the solvent (water), and m is the molality of the solute particles in the solution.
We know that the freezing point depression (ΔTf) is -0.56°C, and the Kf for water is 1.86°C/m.
-0.56 = 1.86 * m
Solving for m:
m = -0.56/1.86 = -0.301
However, molality cannot be negative, so we need to take the absolute value of m:
m = 0.301 mol/kg
Therefore, the osmolarity of the biological fluid is 0.301 osmol/kg.
To answer this question, we need to use the freezing point depression equation:
ΔTf = Kf * m
Where ΔTf is the change in freezing point, Kf is the freezing point depression constant for the solvent (water), and m is the molality of the solute particles in the solution.
We know that the freezing point depression (ΔTf) is -0.56°C, and the Kf for water is 1.86°C/m.
-0.56 = 1.86 * m
Solving for m:
m = -0.56/1.86 = -0.301
However, molality cannot be negative, so we need to take the absolute value of m:
m = 0.301 mol/kg
Therefore, the osmolarity of the biological fluid is 0.301 osmol/kg.
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Freezing point of a biological fluid is - 0.56° C in aqueous solution.Kf (H20) = 1 .86° mol-1 kg. Thus, its osmotic pressure at 310 K is (assume molarity = molality)a)0.0766 atmb)7.66 atmc)0.766 atmd)0.914 atmCorrect answer is option 'B'. Can you explain this answer?
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Freezing point of a biological fluid is - 0.56° C in aqueous solution.Kf (H20) = 1 .86° mol-1 kg. Thus, its osmotic pressure at 310 K is (assume molarity = molality)a)0.0766 atmb)7.66 atmc)0.766 atmd)0.914 atmCorrect answer is option 'B'. 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 Freezing point of a biological fluid is - 0.56° C in aqueous solution.Kf (H20) = 1 .86° mol-1 kg. Thus, its osmotic pressure at 310 K is (assume molarity = molality)a)0.0766 atmb)7.66 atmc)0.766 atmd)0.914 atmCorrect answer is option 'B'. 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 Freezing point of a biological fluid is - 0.56° C in aqueous solution.Kf (H20) = 1 .86° mol-1 kg. Thus, its osmotic pressure at 310 K is (assume molarity = molality)a)0.0766 atmb)7.66 atmc)0.766 atmd)0.914 atmCorrect answer is option 'B'. Can you explain this answer?.
Freezing point of a biological fluid is - 0.56° C in aqueous solution.Kf (H20) = 1 .86° mol-1 kg. Thus, its osmotic pressure at 310 K is (assume molarity = molality)a)0.0766 atmb)7.66 atmc)0.766 atmd)0.914 atmCorrect answer is option 'B'. 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 Freezing point of a biological fluid is - 0.56° C in aqueous solution.Kf (H20) = 1 .86° mol-1 kg. Thus, its osmotic pressure at 310 K is (assume molarity = molality)a)0.0766 atmb)7.66 atmc)0.766 atmd)0.914 atmCorrect answer is option 'B'. 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 Freezing point of a biological fluid is - 0.56° C in aqueous solution.Kf (H20) = 1 .86° mol-1 kg. Thus, its osmotic pressure at 310 K is (assume molarity = molality)a)0.0766 atmb)7.66 atmc)0.766 atmd)0.914 atmCorrect answer is option 'B'. Can you explain this answer?.
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