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Glucose solution is one molal. Glucose present in 1 kg glucose solution is
- a)180 g
- b)153 g
- c)820 g
- d)1 mole
Correct answer is 'B'. Can you explain this answer?
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Verified Answer
Glucose solution is one molal. Glucose present in 1 kg glucose solutio...
Glucose solution is one molal. Thus, 1000 g water (solvent) has glucose = one mole = 180 g
Total mass of solution = 1000 +180 = 1180 g
1180 g of solution has glucose = 180 g
∴ 1000 g of solution has glucose =
180 x 1000/1180 = 152.5 ≈ 153g
180 x 1000/1180 = 152.5 ≈ 153g
Most Upvoted Answer
Glucose solution is one molal. Glucose present in 1 kg glucose solutio...
**Explanation:**
To find the amount of glucose present in the 1 kg glucose solution, we need to first understand the concept of molality.
Molality (m) is a measure of the concentration of a solute in a solution, expressed as the number of moles of solute per kilogram of solvent. In this case, the solute is glucose and the solvent is water.
Given that the solution is one molal, it means that there is one mole of glucose dissolved in one kilogram of water.
**Calculating the amount of glucose:**
To calculate the amount of glucose present in the 1 kg glucose solution, we need to convert the molality into moles of glucose.
Given:
Molality (m) = 1 mol/kg
Mass of solvent (water) = 1 kg
To find the moles of glucose, we can use the formula:
moles of solute = molality * mass of solvent
moles of glucose = 1 mol/kg * 1 kg = 1 mol
Therefore, the amount of glucose present in the 1 kg glucose solution is 1 mole.
**Converting moles to grams:**
To find the amount of glucose in grams, we need to use the molecular weight of glucose, which is 180 g/mol.
Amount of glucose = moles of glucose * molecular weight of glucose
Amount of glucose = 1 mol * 180 g/mol = 180 g
Hence, the correct answer is option (a) 180 g.
To find the amount of glucose present in the 1 kg glucose solution, we need to first understand the concept of molality.
Molality (m) is a measure of the concentration of a solute in a solution, expressed as the number of moles of solute per kilogram of solvent. In this case, the solute is glucose and the solvent is water.
Given that the solution is one molal, it means that there is one mole of glucose dissolved in one kilogram of water.
**Calculating the amount of glucose:**
To calculate the amount of glucose present in the 1 kg glucose solution, we need to convert the molality into moles of glucose.
Given:
Molality (m) = 1 mol/kg
Mass of solvent (water) = 1 kg
To find the moles of glucose, we can use the formula:
moles of solute = molality * mass of solvent
moles of glucose = 1 mol/kg * 1 kg = 1 mol
Therefore, the amount of glucose present in the 1 kg glucose solution is 1 mole.
**Converting moles to grams:**
To find the amount of glucose in grams, we need to use the molecular weight of glucose, which is 180 g/mol.
Amount of glucose = moles of glucose * molecular weight of glucose
Amount of glucose = 1 mol * 180 g/mol = 180 g
Hence, the correct answer is option (a) 180 g.
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Glucose solution is one molal. Glucose present in 1 kg glucose solution isa)180 gb)153 gc)820 gd)1 moleCorrect answer is '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 Glucose solution is one molal. Glucose present in 1 kg glucose solution isa)180 gb)153 gc)820 gd)1 moleCorrect answer is '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 Glucose solution is one molal. Glucose present in 1 kg glucose solution isa)180 gb)153 gc)820 gd)1 moleCorrect answer is 'B'. Can you explain this answer?.
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