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100 g of C6H12O6 (aq.) solution has vapour pressure is equal to 40 torr at certain temperature. Vapour pressure of H2O(l) is 40.18 torr at same temperature. If this solution is cooled to - 0.93 deg C .what mass of ice will be separated out? (Kf) = 1.86kgmol-¹?
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100 g of C6H12O6 (aq.) solution has vapour pressure is equal to 40 tor...
Given data:
Mass of C6H12O6 (aq.) solution = 100 g
Vapour pressure of C6H12O6 (aq.) solution = 40 torr
Vapour pressure of H2O(l) = 40.18 torr
Freezing point depression constant (Kf) = 1.86 kgmol-¹
Temperature at which solution is cooled = -0.93°C
To find:
Mass of ice separated out
Solution:
Determine molality of C6H12O6 (aq.) solution
Molality (m) = moles of solute / mass of solvent (kg)
Molar mass of C6H12O6 = 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol
Number of moles of C6H12O6 = mass / molar mass = 100 / 180.18 = 0.555 mol
Mass of H2O in solution = 100 - 0.555 x 180.18 = 9.98 g
Molality of C6H12O6 (aq.) solution = 0.555 / 0.00998 = 55.62 mol/kg
Determine change in freezing point
ΔTf = Kf x m
ΔTf = 1.86 x 55.62 = 103.59°C
Change in freezing point = -0.93 - (-103.59) = 102.66°C
Determine mass of ice separated out
ΔTf = Kf x m x i
Where i is the van't Hoff factor, which is 1 for C6H12O6
Mass of ice separated out = (mass of solvent) x (ΔTf / Kf)
Mass of solvent = 9.98 g
Mass of ice separated out = 9.98 x (102.66 / 1.86) = 545.25 g
Therefore, the mass of ice separated out is 545.25 g.
Mass of C6H12O6 (aq.) solution = 100 g
Vapour pressure of C6H12O6 (aq.) solution = 40 torr
Vapour pressure of H2O(l) = 40.18 torr
Freezing point depression constant (Kf) = 1.86 kgmol-¹
Temperature at which solution is cooled = -0.93°C
To find:
Mass of ice separated out
Solution:
Determine molality of C6H12O6 (aq.) solution
Molality (m) = moles of solute / mass of solvent (kg)
Molar mass of C6H12O6 = 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol
Number of moles of C6H12O6 = mass / molar mass = 100 / 180.18 = 0.555 mol
Mass of H2O in solution = 100 - 0.555 x 180.18 = 9.98 g
Molality of C6H12O6 (aq.) solution = 0.555 / 0.00998 = 55.62 mol/kg
Determine change in freezing point
ΔTf = Kf x m
ΔTf = 1.86 x 55.62 = 103.59°C
Change in freezing point = -0.93 - (-103.59) = 102.66°C
Determine mass of ice separated out
ΔTf = Kf x m x i
Where i is the van't Hoff factor, which is 1 for C6H12O6
Mass of ice separated out = (mass of solvent) x (ΔTf / Kf)
Mass of solvent = 9.98 g
Mass of ice separated out = 9.98 x (102.66 / 1.86) = 545.25 g
Therefore, the mass of ice separated out is 545.25 g.
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100 g of C6H12O6 (aq.) solution has vapour pressure is equal to 40 torr at certain temperature. Vapour pressure of H2O(l) is 40.18 torr at same temperature. If this solution is cooled to - 0.93 deg C .what mass of ice will be separated out? (Kf) = 1.86kgmol-¹?
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100 g of C6H12O6 (aq.) solution has vapour pressure is equal to 40 torr at certain temperature. Vapour pressure of H2O(l) is 40.18 torr at same temperature. If this solution is cooled to - 0.93 deg C .what mass of ice will be separated out? (Kf) = 1.86kgmol-¹? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 100 g of C6H12O6 (aq.) solution has vapour pressure is equal to 40 torr at certain temperature. Vapour pressure of H2O(l) is 40.18 torr at same temperature. If this solution is cooled to - 0.93 deg C .what mass of ice will be separated out? (Kf) = 1.86kgmol-¹? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 100 g of C6H12O6 (aq.) solution has vapour pressure is equal to 40 torr at certain temperature. Vapour pressure of H2O(l) is 40.18 torr at same temperature. If this solution is cooled to - 0.93 deg C .what mass of ice will be separated out? (Kf) = 1.86kgmol-¹?.
100 g of C6H12O6 (aq.) solution has vapour pressure is equal to 40 torr at certain temperature. Vapour pressure of H2O(l) is 40.18 torr at same temperature. If this solution is cooled to - 0.93 deg C .what mass of ice will be separated out? (Kf) = 1.86kgmol-¹? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 100 g of C6H12O6 (aq.) solution has vapour pressure is equal to 40 torr at certain temperature. Vapour pressure of H2O(l) is 40.18 torr at same temperature. If this solution is cooled to - 0.93 deg C .what mass of ice will be separated out? (Kf) = 1.86kgmol-¹? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 100 g of C6H12O6 (aq.) solution has vapour pressure is equal to 40 torr at certain temperature. Vapour pressure of H2O(l) is 40.18 torr at same temperature. If this solution is cooled to - 0.93 deg C .what mass of ice will be separated out? (Kf) = 1.86kgmol-¹?.
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100 g of C6H12O6 (aq.) solution has vapour pressure is equal to 40 torr at certain temperature. Vapour pressure of H2O(l) is 40.18 torr at same temperature. If this solution is cooled to - 0.93 deg C .what mass of ice will be separated out? (Kf) = 1.86kgmol-¹?, a detailed solution for 100 g of C6H12O6 (aq.) solution has vapour pressure is equal to 40 torr at certain temperature. Vapour pressure of H2O(l) is 40.18 torr at same temperature. If this solution is cooled to - 0.93 deg C .what mass of ice will be separated out? (Kf) = 1.86kgmol-¹? has been provided alongside types of 100 g of C6H12O6 (aq.) solution has vapour pressure is equal to 40 torr at certain temperature. Vapour pressure of H2O(l) is 40.18 torr at same temperature. If this solution is cooled to - 0.93 deg C .what mass of ice will be separated out? (Kf) = 1.86kgmol-¹? theory, EduRev gives you an
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