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The apparent degree of ionization of KCl in water at 290 K is 0.86, the mass of KCl which must be made up to 1 dm3 of aqueous solution so as to produce the same osmotic pressure as the 4% solution of glucose at that temperature is ______g.
Correct answer is '8.9'. Can you explain this answer?
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The apparent degree of ionization of KCl in water at 290 K is 0.86, th...
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The apparent degree of ionization of KCl in water at 290 K is 0.86, th...
Calculation of Van't Hoff factor:
The apparent degree of ionization of KCl in water is 0.86. Therefore, the Van't Hoff factor (i) can be calculated as follows:
i = 1 + α(n-1)
where α is the degree of ionization, n is the number of ions produced per molecule in solution.
For KCl, n = 2 (one K+ ion and one Cl- ion are produced per KCl molecule in solution).
Therefore, i = 1 + 0.86(2-1) = 1.86
Calculation of molality of glucose solution:
The molality (m) of a solution is defined as the number of moles of solute per kilogram of solvent. A 4% solution of glucose means that 4 g of glucose is dissolved in 100 ml of solution.
Therefore, the molality of the glucose solution can be calculated as follows:
Mass of glucose in 1 L of solution = (4/100) x 1000 g = 40 g
Molar mass of glucose = 180.16 g/mol
Number of moles of glucose = 40/180.16 = 0.222 mol
Mass of water in 1 L of solution = 1000 g - 40 g = 960 g
Therefore, the molality of the glucose solution = 0.222 mol/0.96 kg = 0.23 mol/kg
Calculation of mass of KCl:
The osmotic pressure (π) of a solution is directly proportional to the molality of the solute. Therefore, we can use the following equation to calculate the mass of KCl required to produce the same osmotic pressure as the glucose solution:
π = i m R T
where R is the gas constant and T is the temperature in Kelvin.
For water at 290 K, R = 0.0821 L atm/mol K.
Substituting the values of i, m, R, and T, we get:
π = 1.86 x 0.23 x 0.0821 x 290 = 10.7 atm
The osmotic pressure of the 4% glucose solution is also 10.7 atm.
The mass of KCl required can be calculated as follows:
Molarity of KCl solution = molality of KCl solution / molar mass of KCl
Molality of KCl solution = molality of glucose solution / i
Therefore, molarity of KCl solution = 0.23 mol/kg / 1.86 = 0.124 mol/L
Mass of KCl required to make 1 L of solution = molarity of KCl solution x molar mass of KCl
= 0.124 mol/L x 74.55 g/mol = 9.26 g/L
Therefore, the mass of KCl required to make 1 dm3 (1000 ml) of solution = 9.26 g/L x 1000 ml / 1000 = 8.9 g.
The apparent degree of ionization of KCl in water is 0.86. Therefore, the Van't Hoff factor (i) can be calculated as follows:
i = 1 + α(n-1)
where α is the degree of ionization, n is the number of ions produced per molecule in solution.
For KCl, n = 2 (one K+ ion and one Cl- ion are produced per KCl molecule in solution).
Therefore, i = 1 + 0.86(2-1) = 1.86
Calculation of molality of glucose solution:
The molality (m) of a solution is defined as the number of moles of solute per kilogram of solvent. A 4% solution of glucose means that 4 g of glucose is dissolved in 100 ml of solution.
Therefore, the molality of the glucose solution can be calculated as follows:
Mass of glucose in 1 L of solution = (4/100) x 1000 g = 40 g
Molar mass of glucose = 180.16 g/mol
Number of moles of glucose = 40/180.16 = 0.222 mol
Mass of water in 1 L of solution = 1000 g - 40 g = 960 g
Therefore, the molality of the glucose solution = 0.222 mol/0.96 kg = 0.23 mol/kg
Calculation of mass of KCl:
The osmotic pressure (π) of a solution is directly proportional to the molality of the solute. Therefore, we can use the following equation to calculate the mass of KCl required to produce the same osmotic pressure as the glucose solution:
π = i m R T
where R is the gas constant and T is the temperature in Kelvin.
For water at 290 K, R = 0.0821 L atm/mol K.
Substituting the values of i, m, R, and T, we get:
π = 1.86 x 0.23 x 0.0821 x 290 = 10.7 atm
The osmotic pressure of the 4% glucose solution is also 10.7 atm.
The mass of KCl required can be calculated as follows:
Molarity of KCl solution = molality of KCl solution / molar mass of KCl
Molality of KCl solution = molality of glucose solution / i
Therefore, molarity of KCl solution = 0.23 mol/kg / 1.86 = 0.124 mol/L
Mass of KCl required to make 1 L of solution = molarity of KCl solution x molar mass of KCl
= 0.124 mol/L x 74.55 g/mol = 9.26 g/L
Therefore, the mass of KCl required to make 1 dm3 (1000 ml) of solution = 9.26 g/L x 1000 ml / 1000 = 8.9 g.
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The apparent degree of ionization of KCl in water at 290 K is 0.86, the mass of KCl which must be made up to 1 dm3 of aqueous solution so as to produce the same osmotic pressure as the 4% solution of glucose at that temperature is ______g.Correct answer is '8.9'. Can you explain this answer?
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The apparent degree of ionization of KCl in water at 290 K is 0.86, the mass of KCl which must be made up to 1 dm3 of aqueous solution so as to produce the same osmotic pressure as the 4% solution of glucose at that temperature is ______g.Correct answer is '8.9'. 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 The apparent degree of ionization of KCl in water at 290 K is 0.86, the mass of KCl which must be made up to 1 dm3 of aqueous solution so as to produce the same osmotic pressure as the 4% solution of glucose at that temperature is ______g.Correct answer is '8.9'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The apparent degree of ionization of KCl in water at 290 K is 0.86, the mass of KCl which must be made up to 1 dm3 of aqueous solution so as to produce the same osmotic pressure as the 4% solution of glucose at that temperature is ______g.Correct answer is '8.9'. Can you explain this answer?.
The apparent degree of ionization of KCl in water at 290 K is 0.86, the mass of KCl which must be made up to 1 dm3 of aqueous solution so as to produce the same osmotic pressure as the 4% solution of glucose at that temperature is ______g.Correct answer is '8.9'. 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 The apparent degree of ionization of KCl in water at 290 K is 0.86, the mass of KCl which must be made up to 1 dm3 of aqueous solution so as to produce the same osmotic pressure as the 4% solution of glucose at that temperature is ______g.Correct answer is '8.9'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The apparent degree of ionization of KCl in water at 290 K is 0.86, the mass of KCl which must be made up to 1 dm3 of aqueous solution so as to produce the same osmotic pressure as the 4% solution of glucose at that temperature is ______g.Correct answer is '8.9'. Can you explain this answer?.
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