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An aqueous solution of glucose boils at 100.01ºC. The mo lal elevat ion constant for water is 0.5 K Kg mol–1. The number of glucose molecules in the solution containing 100 g of water are x × 1021. The value of x is ______.
Correct answer is '1.2'. Can you explain this answer?
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An aqueous solution of glucose boils at 100.01ºC. The mo lal elev...
Solution:
Given, boiling point elevation = 0.01C
Molal elevation constant, Kf = 0.5 K Kg mol-1
Mass of water, w2 = 100 g
Number of glucose molecules in solution = x x 1021
We know that the boiling point elevation is given by the equation:
ΔTb = Kf × m
where ΔTb is the boiling point elevation, Kf is the molal elevation constant and m is the molality of the solution.
To find the molality of the solution, we need to know the mass of glucose in the solution and the mass of water in the solution.
Let the mass of glucose in the solution be w1.
Then, the total mass of the solution is given by:
w = w1 + w2
The molality of the solution is given by:
m = nglucose / mwater
where nglucose is the number of moles of glucose and mwater is the mass of water in kg.
The number of moles of glucose is given by:
nglucose = w1 / Mglucose
where Mglucose is the molar mass of glucose.
Substituting the values, we get:
ΔTb = Kf × nglucose / mwater
Solving for nglucose, we get:
nglucose = ΔTb × mwater / Kf
Substituting the given values, we get:
nglucose = 0.01 × 100 / 0.5
nglucose = 2
The number of moles of glucose in the solution is 2.
The number of glucose molecules in the solution is given by:
nglucose = N / Avogadro's number
where N is the number of glucose molecules and Avogadro's number is 6.022 x 1023 mol-1.
Substituting the values, we get:
N = nglucose × Avogadro's number
N = 2 × 6.022 x 1023
N = 1.2044 x 1024
The number of glucose molecules in the solution is 1.2044 x 1024.
Rounding off to one significant figure, we get:
x = 1.2
Therefore, the value of x is 1.2.
Given, boiling point elevation = 0.01C
Molal elevation constant, Kf = 0.5 K Kg mol-1
Mass of water, w2 = 100 g
Number of glucose molecules in solution = x x 1021
We know that the boiling point elevation is given by the equation:
ΔTb = Kf × m
where ΔTb is the boiling point elevation, Kf is the molal elevation constant and m is the molality of the solution.
To find the molality of the solution, we need to know the mass of glucose in the solution and the mass of water in the solution.
Let the mass of glucose in the solution be w1.
Then, the total mass of the solution is given by:
w = w1 + w2
The molality of the solution is given by:
m = nglucose / mwater
where nglucose is the number of moles of glucose and mwater is the mass of water in kg.
The number of moles of glucose is given by:
nglucose = w1 / Mglucose
where Mglucose is the molar mass of glucose.
Substituting the values, we get:
ΔTb = Kf × nglucose / mwater
Solving for nglucose, we get:
nglucose = ΔTb × mwater / Kf
Substituting the given values, we get:
nglucose = 0.01 × 100 / 0.5
nglucose = 2
The number of moles of glucose in the solution is 2.
The number of glucose molecules in the solution is given by:
nglucose = N / Avogadro's number
where N is the number of glucose molecules and Avogadro's number is 6.022 x 1023 mol-1.
Substituting the values, we get:
N = nglucose × Avogadro's number
N = 2 × 6.022 x 1023
N = 1.2044 x 1024
The number of glucose molecules in the solution is 1.2044 x 1024.
Rounding off to one significant figure, we get:
x = 1.2
Therefore, the value of x is 1.2.
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An aqueous solution of glucose boils at 100.01ºC. The mo lal elevat ion constant for water is 0.5 K Kg mol–1. The number of glucose molecules in the solution containing 100 g of water are x × 1021. The value of x is ______.Correct answer is '1.2'. Can you explain this answer?
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An aqueous solution of glucose boils at 100.01ºC. The mo lal elevat ion constant for water is 0.5 K Kg mol–1. The number of glucose molecules in the solution containing 100 g of water are x × 1021. The value of x is ______.Correct answer is '1.2'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about An aqueous solution of glucose boils at 100.01ºC. The mo lal elevat ion constant for water is 0.5 K Kg mol–1. The number of glucose molecules in the solution containing 100 g of water are x × 1021. The value of x is ______.Correct answer is '1.2'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An aqueous solution of glucose boils at 100.01ºC. The mo lal elevat ion constant for water is 0.5 K Kg mol–1. The number of glucose molecules in the solution containing 100 g of water are x × 1021. The value of x is ______.Correct answer is '1.2'. Can you explain this answer?.
Solutions for An aqueous solution of glucose boils at 100.01ºC. The mo lal elevat ion constant for water is 0.5 K Kg mol–1. The number of glucose molecules in the solution containing 100 g of water are x × 1021. The value of x is ______.Correct answer is '1.2'. Can you explain this answer? in English & in Hindi are available as part of our courses for Chemistry.
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