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At 27° C, an aqueous solution of a carbohydrate containing 18 g per 100 mL of the solution is found to be isotonic with aqueous solution containing 2.925 g NaCI per 100 mL of the solution. Thus, molar mass of the carbohydrate is (assume NaCI is completely ionised)
- a)360.0 g mol-1
- b)180.0 g mol-1
- c)90.0 g mol-1
- d)58.5 g mol-1
Correct answer is option 'B'. Can you explain this answer?
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At 27 C, an aqueous solution of a carbohydrate containing 18 g per 100...
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At 27 C, an aqueous solution of a carbohydrate containing 18 g per 100...
Given:
- Temperature (T) = 27°C
- Concentration of carbohydrate solution (C₁) = 18 g/100 mL
- Concentration of NaCl solution (C₂) = 2.925 g/100 mL
To find:
Molar mass of the carbohydrate
Explanation:
1. Isotonic Solution:
An isotonic solution is a solution that has the same osmotic pressure as another solution. In this case, the aqueous solution of the carbohydrate is isotonic with the NaCl solution.
2. Osmotic Pressure:
Osmotic pressure (π) is the pressure required to prevent the osmosis of a solvent across a semipermeable membrane. It depends on the concentration of solute particles in the solution.
3. Osmotic Pressure Formula:
The osmotic pressure (π) of a solution is given by the equation:
π = nRT/V
Where,
- n is the number of moles of solute particles
- R is the ideal gas constant (0.0821 L·atm/(mol·K))
- T is the temperature in Kelvin (K)
- V is the volume of the solution in liters (L)
4. Calculation of Osmotic Pressure for Carbohydrate Solution:
First, we need to calculate the osmotic pressure of the carbohydrate solution using the given concentration and temperature.
Concentration of carbohydrate solution (C₁) = 18 g/100 mL
Molar mass of carbohydrate (M) = ?
We can assume that the carbohydrate is non-ionizing, so the number of moles of solute particles (n₁) is equal to the number of moles of the carbohydrate (n).
Given concentration is in grams per 100 mL, so we need to convert it to moles per liter (mol/L) for the osmotic pressure equation.
Concentration in mol/L (C₁') = (C₁ * 10) / M
Now, we can substitute the values in the osmotic pressure equation:
π₁ = n₁RT/V = (C₁' * RT)
5. Calculation of Osmotic Pressure for NaCl Solution:
Similarly, we need to calculate the osmotic pressure of the NaCl solution using the given concentration and temperature.
Concentration of NaCl solution (C₂) = 2.925 g/100 mL
Molar mass of NaCl (Mₚ) = 58.5 g/mol
Since NaCl is completely ionized, the number of moles of solute particles (n₂) is twice the number of moles of NaCl (nₚ).
Concentration in mol/L (C₂') = (C₂ * 10) / Mₚ
Now, we can substitute the values in the osmotic pressure equation:
π₂ = n₂RT/V = (2 * C₂' * RT)
6. Equating Osmotic Pressures:
Since the two solutions are isotonic, their osmotic pressures should be equal.
π₁ = π₂
(C₁' * RT) = (2 * C₂' * RT)
7. Solving for the Molar Mass of the Carbohydrate:
- Temperature (T) = 27°C
- Concentration of carbohydrate solution (C₁) = 18 g/100 mL
- Concentration of NaCl solution (C₂) = 2.925 g/100 mL
To find:
Molar mass of the carbohydrate
Explanation:
1. Isotonic Solution:
An isotonic solution is a solution that has the same osmotic pressure as another solution. In this case, the aqueous solution of the carbohydrate is isotonic with the NaCl solution.
2. Osmotic Pressure:
Osmotic pressure (π) is the pressure required to prevent the osmosis of a solvent across a semipermeable membrane. It depends on the concentration of solute particles in the solution.
3. Osmotic Pressure Formula:
The osmotic pressure (π) of a solution is given by the equation:
π = nRT/V
Where,
- n is the number of moles of solute particles
- R is the ideal gas constant (0.0821 L·atm/(mol·K))
- T is the temperature in Kelvin (K)
- V is the volume of the solution in liters (L)
4. Calculation of Osmotic Pressure for Carbohydrate Solution:
First, we need to calculate the osmotic pressure of the carbohydrate solution using the given concentration and temperature.
Concentration of carbohydrate solution (C₁) = 18 g/100 mL
Molar mass of carbohydrate (M) = ?
We can assume that the carbohydrate is non-ionizing, so the number of moles of solute particles (n₁) is equal to the number of moles of the carbohydrate (n).
Given concentration is in grams per 100 mL, so we need to convert it to moles per liter (mol/L) for the osmotic pressure equation.
Concentration in mol/L (C₁') = (C₁ * 10) / M
Now, we can substitute the values in the osmotic pressure equation:
π₁ = n₁RT/V = (C₁' * RT)
5. Calculation of Osmotic Pressure for NaCl Solution:
Similarly, we need to calculate the osmotic pressure of the NaCl solution using the given concentration and temperature.
Concentration of NaCl solution (C₂) = 2.925 g/100 mL
Molar mass of NaCl (Mₚ) = 58.5 g/mol
Since NaCl is completely ionized, the number of moles of solute particles (n₂) is twice the number of moles of NaCl (nₚ).
Concentration in mol/L (C₂') = (C₂ * 10) / Mₚ
Now, we can substitute the values in the osmotic pressure equation:
π₂ = n₂RT/V = (2 * C₂' * RT)
6. Equating Osmotic Pressures:
Since the two solutions are isotonic, their osmotic pressures should be equal.
π₁ = π₂
(C₁' * RT) = (2 * C₂' * RT)
7. Solving for the Molar Mass of the Carbohydrate:
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At 27 C, an aqueous solution of a carbohydrate containing 18 g per 100 mL of the solution is found to be isotonic with aqueous solution containing 2.925 g NaCI per 100 mL of the solution. Thus, molar mass of the carbohydrate is (assume NaCI is completely ionised)a)360.0 g mol-1b)180.0g mol-1c)90.0g mol-1d)58.5g mol-1Correct answer is option 'B'. Can you explain this answer?
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At 27 C, an aqueous solution of a carbohydrate containing 18 g per 100 mL of the solution is found to be isotonic with aqueous solution containing 2.925 g NaCI per 100 mL of the solution. Thus, molar mass of the carbohydrate is (assume NaCI is completely ionised)a)360.0 g mol-1b)180.0g mol-1c)90.0g mol-1d)58.5g mol-1Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about At 27 C, an aqueous solution of a carbohydrate containing 18 g per 100 mL of the solution is found to be isotonic with aqueous solution containing 2.925 g NaCI per 100 mL of the solution. Thus, molar mass of the carbohydrate is (assume NaCI is completely ionised)a)360.0 g mol-1b)180.0g mol-1c)90.0g mol-1d)58.5g mol-1Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At 27 C, an aqueous solution of a carbohydrate containing 18 g per 100 mL of the solution is found to be isotonic with aqueous solution containing 2.925 g NaCI per 100 mL of the solution. Thus, molar mass of the carbohydrate is (assume NaCI is completely ionised)a)360.0 g mol-1b)180.0g mol-1c)90.0g mol-1d)58.5g mol-1Correct answer is option 'B'. Can you explain this answer?.
At 27 C, an aqueous solution of a carbohydrate containing 18 g per 100 mL of the solution is found to be isotonic with aqueous solution containing 2.925 g NaCI per 100 mL of the solution. Thus, molar mass of the carbohydrate is (assume NaCI is completely ionised)a)360.0 g mol-1b)180.0g mol-1c)90.0g mol-1d)58.5g mol-1Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about At 27 C, an aqueous solution of a carbohydrate containing 18 g per 100 mL of the solution is found to be isotonic with aqueous solution containing 2.925 g NaCI per 100 mL of the solution. Thus, molar mass of the carbohydrate is (assume NaCI is completely ionised)a)360.0 g mol-1b)180.0g mol-1c)90.0g mol-1d)58.5g mol-1Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At 27 C, an aqueous solution of a carbohydrate containing 18 g per 100 mL of the solution is found to be isotonic with aqueous solution containing 2.925 g NaCI per 100 mL of the solution. Thus, molar mass of the carbohydrate is (assume NaCI is completely ionised)a)360.0 g mol-1b)180.0g mol-1c)90.0g mol-1d)58.5g mol-1Correct answer is option 'B'. Can you explain this answer?.
Solutions for At 27 C, an aqueous solution of a carbohydrate containing 18 g per 100 mL of the solution is found to be isotonic with aqueous solution containing 2.925 g NaCI per 100 mL of the solution. Thus, molar mass of the carbohydrate is (assume NaCI is completely ionised)a)360.0 g mol-1b)180.0g mol-1c)90.0g mol-1d)58.5g mol-1Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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ample number of questions to practice At 27 C, an aqueous solution of a carbohydrate containing 18 g per 100 mL of the solution is found to be isotonic with aqueous solution containing 2.925 g NaCI per 100 mL of the solution. Thus, molar mass of the carbohydrate is (assume NaCI is completely ionised)a)360.0 g mol-1b)180.0g mol-1c)90.0g mol-1d)58.5g mol-1Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.
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