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At 300 K, 36 g of glucose present per litre in its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?
At 300 K, 36 g of glucose present per litre in its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?
0.061 M
18 g of glucose, $C_6H_{12}O_6$, (Molar Mass= 180 g mol$^{-1}$) is dissolved in 1 kg of water in a sauce pan. At what temperature will this solution boil (Kb for water = 0.52 K kg mol$^{-1}$)?
18 g of glucose, $C_6H_{12}O_6$, (Molar Mass= 180 g mol$^{-1}$) is dissolved in 1 kg of water in a sauce pan. At what temperature will this solution boil (Kb for water = 0.52 K kg mol$^{-1}$)?
100.052 $^\circ$C
Calculate the boiling point of a solution prepared by adding 15.00 g of NaCl to 250.0g of water. ($K_b$ for water = 0.512 K kg mol$^{-1}$, (Molar mass of NaCl = 58.44 g)).
Calculate the boiling point of a solution prepared by adding 15.00 g of NaCl to 250.0g of water. ($K_b$ for water = 0.512 K kg mol$^{-1}$, (Molar mass of NaCl = 58.44 g)).
101.5 $^\circ$C
Flashcards
Boiling Point Elevation
Boiling Point Elevation
The increase in boiling point of a solution compared to the pure solvent.
Colligative Property
Colligative Property
A colligative property dependent on the number of solute particles in a solution.
Van't Hoff Factor (i)
Van't Hoff Factor (i)
Van't Hoff factor represents the number of particles a solute dissociates into in solution.
ΔTb Formula
ΔTb Formula
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Kb
Kb
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W2
W2
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W1
W1
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M2
M2
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NaCl Dissociation
NaCl Dissociation
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Solution Boiling Point
Solution Boiling Point
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Osmotic Pressure (π)
Osmotic Pressure (π)
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van't Hoff Equation
van't Hoff Equation
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Molar concentration
Molar concentration
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R
R
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T
T
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Molar Mass
Molar Mass
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Solution
Solution
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Solute
Solute
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Solvent
Solvent
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Freezing Point Depression
Freezing Point Depression
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Study Notes
- Problem: Calculating the boiling point of a solution made by adding 15.00 g of NaCl to 250.0 g of water.
- Water's ebullioscopic constant (Kb) is 0.512 K kg mol-1.
- NaCl's molar mass is 58.44 g.
Calculation Steps
- ΔTb = (i * Kb * 1000 * W2) / (W1 * M2)
- NaCl dissociates into Na+ and Cl- ions, resulting in i = 2.
- W2 = 15.0 g (mass of NaCl)
- W1 = 250.0 g (mass of water)
- M2 = 58.44 g mol-1 (molar mass of NaCl)
- Kb = 0.512 K kg mol-1
- ΔTb = (2 * 0.512 * 1000 * 15.0) / (250.0 * 58.44) = 1.05 °C.
- Boiling point of the solution = 100 + 1.05 = 101.5 °C.
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Description
Calculate the boiling point of a NaCl solution. The solution is made by adding 15.00 g of NaCl to 250.0 g of water. The boiling point elevation is calculated using the formula ΔTb = (i * Kb * 1000 * W2) / (W1 * M2).
