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The mass of glucose that would be dissolved in 50g of water in order to produce the same lowering of vapour pressure as is produced by dissolving 1 g of Urea in the same quantity of water is?
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The mass of glucose that would be dissolved in 50g of water in order t...
1 g of urea is 60 times less than its MW. Mass of glucose required should also be 60 times less, that is 180g /60 = 3 g
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The mass of glucose that would be dissolved in 50g of water in order t...
Calculation of Mass of Glucose to produce the same lowering of vapour pressure
In order to calculate the mass of glucose that would be dissolved in 50g of water to produce the same lowering of vapour pressure as 1g of Urea in the same quantity of water, we can use the formula:
$\Delta P = i \cdot K_f \cdot m$
Where:
- $\Delta P$ = lowering of vapour pressure
- i = Van't Hoff factor (1 for glucose and 1 for urea)
- $K_f$ = cryoscopic constant for water (1.86 °C kg/mol)
- m = molality of the solution
Calculating Molality of Urea solution
Given that 1g of Urea is dissolved in 50g of water, we first need to calculate the molality of the Urea solution:
Molar mass of Urea (CH4N2O) = 60 g/mol
Number of moles of Urea = 1g / 60 g/mol = 0.0167 mol
Mass of water = 50g
Molality = moles of solute / mass of solvent (in kg) = 0.0167 mol / 0.05 kg = 0.334 mol/kg
Calculating Mass of Glucose
Now, we can use the molality of the Urea solution to calculate the mass of glucose that would produce the same lowering of vapour pressure:
$\Delta P_{Urea} = \Delta P_{Glucose}$
$i \cdot K_f \cdot m_{Urea} = i \cdot K_f \cdot m_{Glucose}$
$0.334 \cdot 1.86 = 1 \cdot 1.86 \cdot m_{Glucose}$
$m_{Glucose} = 0.334$
Therefore, the mass of glucose that would need to be dissolved in 50g of water to produce the same lowering of vapour pressure as 1g of Urea is 0.334g.
In order to calculate the mass of glucose that would be dissolved in 50g of water to produce the same lowering of vapour pressure as 1g of Urea in the same quantity of water, we can use the formula:
$\Delta P = i \cdot K_f \cdot m$
Where:
- $\Delta P$ = lowering of vapour pressure
- i = Van't Hoff factor (1 for glucose and 1 for urea)
- $K_f$ = cryoscopic constant for water (1.86 °C kg/mol)
- m = molality of the solution
Calculating Molality of Urea solution
Given that 1g of Urea is dissolved in 50g of water, we first need to calculate the molality of the Urea solution:
Molar mass of Urea (CH4N2O) = 60 g/mol
Number of moles of Urea = 1g / 60 g/mol = 0.0167 mol
Mass of water = 50g
Molality = moles of solute / mass of solvent (in kg) = 0.0167 mol / 0.05 kg = 0.334 mol/kg
Calculating Mass of Glucose
Now, we can use the molality of the Urea solution to calculate the mass of glucose that would produce the same lowering of vapour pressure:
$\Delta P_{Urea} = \Delta P_{Glucose}$
$i \cdot K_f \cdot m_{Urea} = i \cdot K_f \cdot m_{Glucose}$
$0.334 \cdot 1.86 = 1 \cdot 1.86 \cdot m_{Glucose}$
$m_{Glucose} = 0.334$
Therefore, the mass of glucose that would need to be dissolved in 50g of water to produce the same lowering of vapour pressure as 1g of Urea is 0.334g.
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The mass of glucose that would be dissolved in 50g of water in order to produce the same lowering of vapour pressure as is produced by dissolving 1 g of Urea in the same quantity of water is?
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The mass of glucose that would be dissolved in 50g of water in order to produce the same lowering of vapour pressure as is produced by dissolving 1 g of Urea in the same quantity of water is? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The mass of glucose that would be dissolved in 50g of water in order to produce the same lowering of vapour pressure as is produced by dissolving 1 g of Urea in the same quantity of water is? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The mass of glucose that would be dissolved in 50g of water in order to produce the same lowering of vapour pressure as is produced by dissolving 1 g of Urea in the same quantity of water is?.
The mass of glucose that would be dissolved in 50g of water in order to produce the same lowering of vapour pressure as is produced by dissolving 1 g of Urea in the same quantity of water is? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The mass of glucose that would be dissolved in 50g of water in order to produce the same lowering of vapour pressure as is produced by dissolving 1 g of Urea in the same quantity of water is? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The mass of glucose that would be dissolved in 50g of water in order to produce the same lowering of vapour pressure as is produced by dissolving 1 g of Urea in the same quantity of water is?.
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