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18 g of glucose (C6H12O6) is present in 1000 g of an aqueous solution of glucose. The molality of this solution is:
- a)1M
- b)0.1M
- c)1.1M
- d)0.5M
Correct answer is option 'B'. Can you explain this answer?
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18 g of glucose (C6H12O6) is present in 1000 g of an aqueous solution ...
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18 g of glucose (C6H12O6) is present in 1000 g of an aqueous solution ...
Answer: b) 0.1 m.
Explanation:
molality (m) = (1000 × mass of solute) ÷ (molar mass of solute × mass of solvent)
= (1000 × 18) ÷ (180 × 1000)
= 0.1 molal
Explanation:
molality (m) = (1000 × mass of solute) ÷ (molar mass of solute × mass of solvent)
= (1000 × 18) ÷ (180 × 1000)
= 0.1 molal
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18 g of glucose (C6H12O6) is present in 1000 g of an aqueous solution ...
Given data:
Mass of glucose = 18 g
Mass of solution = 1000 g
To find:
Molality of the solution
Formula for molality:
Molality (m) = (moles of solute) / (mass of solvent in kg)
Step 1: Calculate the moles of glucose
Moles of glucose = (mass of glucose) / (molar mass of glucose)
The molar mass of glucose (C6H12O6) can be calculated as follows:
Molar mass of carbon (C) = 12.01 g/mol (atomic mass of carbon)
Molar mass of hydrogen (H) = 1.008 g/mol (atomic mass of hydrogen)
Molar mass of oxygen (O) = 16.00 g/mol (atomic mass of oxygen)
Molar mass of glucose = (6 * molar mass of carbon) + (12 * molar mass of hydrogen) + (6 * molar mass of oxygen)
= (6 * 12.01) + (12 * 1.008) + (6 * 16.00)
= 72.06 + 12.096 + 96.00
= 180.156 g/mol
Moles of glucose = (18 g) / (180.156 g/mol)
= 0.0999 mol (approximately)
Step 2: Calculate the mass of solvent in kg
Mass of solvent = (mass of solution) - (mass of solute)
= 1000 g - 18 g
= 982 g
Mass of solvent in kg = (982 g) / (1000 g/kg)
= 0.982 kg
Step 3: Calculate the molality of the solution
Molality (m) = (moles of solute) / (mass of solvent in kg)
= (0.0999 mol) / (0.982 kg)
= 0.1016 mol/kg (approximately)
Therefore, the molality of the solution is approximately 0.1016 M. However, the given options do not match with this value.
The correct answer is option 'B' (0.1 M), which is the closest value to the calculated molality.
Mass of glucose = 18 g
Mass of solution = 1000 g
To find:
Molality of the solution
Formula for molality:
Molality (m) = (moles of solute) / (mass of solvent in kg)
Step 1: Calculate the moles of glucose
Moles of glucose = (mass of glucose) / (molar mass of glucose)
The molar mass of glucose (C6H12O6) can be calculated as follows:
Molar mass of carbon (C) = 12.01 g/mol (atomic mass of carbon)
Molar mass of hydrogen (H) = 1.008 g/mol (atomic mass of hydrogen)
Molar mass of oxygen (O) = 16.00 g/mol (atomic mass of oxygen)
Molar mass of glucose = (6 * molar mass of carbon) + (12 * molar mass of hydrogen) + (6 * molar mass of oxygen)
= (6 * 12.01) + (12 * 1.008) + (6 * 16.00)
= 72.06 + 12.096 + 96.00
= 180.156 g/mol
Moles of glucose = (18 g) / (180.156 g/mol)
= 0.0999 mol (approximately)
Step 2: Calculate the mass of solvent in kg
Mass of solvent = (mass of solution) - (mass of solute)
= 1000 g - 18 g
= 982 g
Mass of solvent in kg = (982 g) / (1000 g/kg)
= 0.982 kg
Step 3: Calculate the molality of the solution
Molality (m) = (moles of solute) / (mass of solvent in kg)
= (0.0999 mol) / (0.982 kg)
= 0.1016 mol/kg (approximately)
Therefore, the molality of the solution is approximately 0.1016 M. However, the given options do not match with this value.
The correct answer is option 'B' (0.1 M), which is the closest value to the calculated molality.
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18 g of glucose (C6H12O6) is present in 1000 g of an aqueous solution of glucose. The molality of this solution is:a)1Mb)0.1Mc)1.1Md)0.5MCorrect answer is option 'B'. Can you explain this answer?
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