Q. A solution of 0.1 molal urea in water has a freezing point depression of how much? (K_f for water = 1.86 °C kg/mol)
A.
0.186 °C
B.
0.372 °C
C.
0.186 K
D.
0.372 K
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Solution
Freezing point depression = K_f * m = 1.86 * 0.1 = 0.186 °C
Correct Answer: A — 0.186 °C
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Q. Calculate the molality of a solution if the boiling point elevation is 1.024 °C. (K_b for water = 0.512 °C kg/mol)
A.
1 mol/kg
B.
2 mol/kg
C.
0.5 mol/kg
D.
0.25 mol/kg
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Solution
Molality = ΔT_b / (i * K_b) = 1.024 / (2 * 0.512) = 1 mol/kg
Correct Answer: B — 2 mol/kg
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Q. If 0.5 mol of a non-volatile solute is dissolved in 1 kg of water, what is the vapor pressure lowering? (Vapor pressure of pure water = 23.76 mmHg)
A.
1.88 mmHg
B.
2.88 mmHg
C.
3.88 mmHg
D.
4.88 mmHg
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Solution
Vapor pressure lowering = (n_solute / n_solvent) * P°_solvent = (0.5 / 55.5) * 23.76 = 1.88 mmHg
Correct Answer: A — 1.88 mmHg
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Q. If 1 mol of NaCl is dissolved in 1 kg of water, how many particles are present in solution?
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Solution
NaCl dissociates into 2 ions (Na+ and Cl-), so 1 mol of NaCl produces 2 mol of particles in solution.
Correct Answer: B — 2
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Q. If 1 mol of NaCl is dissolved in 1 kg of water, what is the expected van 't Hoff factor (i)?
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Solution
NaCl dissociates into 2 ions (Na+ and Cl-), so the van 't Hoff factor i = 2.
Correct Answer: B — 2
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Q. If 1 mol of NaCl is dissolved in water, how many particles are present in solution?
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Solution
NaCl dissociates into 2 ions (Na+ and Cl-), so 1 mol of NaCl results in 2 mol of particles.
Correct Answer: B — 2
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Q. If 1 mole of a non-electrolyte solute is dissolved in 1 kg of water, what is the freezing point depression?
A.
0 °C
B.
1.86 °C
C.
3.72 °C
D.
5.58 °C
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Solution
The freezing point depression is calculated using the formula ΔTf = i * Kf * m. For a non-electrolyte, i = 1, Kf for water = 1.86 °C kg/mol, and m = 1 mol/kg gives ΔTf = 1.86 °C.
Correct Answer: B — 1.86 °C
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Q. If 1 mole of a non-electrolyte solute is dissolved in 1 kg of water, what is the expected change in freezing point?
A.
0.0 °C
B.
-1.86 °C
C.
-3.72 °C
D.
-5.58 °C
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Solution
The freezing point depression for 1 mole of a non-electrolyte solute in 1 kg of water is -1.86 °C.
Correct Answer: B — -1.86 °C
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Q. If 1 mole of a non-electrolyte solute is dissolved in 1 kg of water, what is the expected freezing point depression?
A.
-1.86 °C
B.
-3.72 °C
C.
-0.52 °C
D.
-2.00 °C
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Solution
The freezing point depression is calculated using the formula ΔTf = Kf * m, where Kf for water is 1.86 °C kg/mol.
Correct Answer: A — -1.86 °C
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Q. If 1 mole of NaCl is dissolved in 1 kg of water, what is the expected van 't Hoff factor (i)?
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Solution
NaCl dissociates into 2 ions (Na+ and Cl-), so the van 't Hoff factor (i) is 2.
Correct Answer: B — 2
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Q. If 2 moles of NaCl are dissolved in 1 kg of water, what is the van 't Hoff factor (i)?
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Solution
NaCl dissociates into 2 ions (Na⁺ and Cl⁻), so the van 't Hoff factor (i) is 2.
Correct Answer: C — 3
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Q. If the osmotic pressure of a solution is 3.0 atm at 25 °C, what is the molarity of the solution? (R = 0.0821 L atm/(K mol))
A.
0.12 M
B.
0.15 M
C.
0.18 M
D.
0.20 M
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Solution
Molarity = π / (RT) = 3.0 / (0.0821 * 298) = 0.12 M
Correct Answer: C — 0.18 M
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Q. If the osmotic pressure of a solution is 3.0 atm, what is the molarity of the solution? (R = 0.0821 L atm/(K mol), T = 298 K)
A.
0.12 M
B.
0.15 M
C.
0.10 M
D.
0.20 M
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Solution
Molarity = π / (RT) = 3.0 / (0.0821 * 298) = 0.12 M
Correct Answer: B — 0.15 M
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Q. What happens to the freezing point of a solvent when a non-volatile solute is added?
A.
It increases
B.
It decreases
C.
It remains the same
D.
It fluctuates
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Solution
The freezing point of a solvent decreases when a non-volatile solute is added, a phenomenon known as freezing point depression.
Correct Answer: B — It decreases
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Q. What happens to the vapor pressure of a solvent when a non-volatile solute is added?
A.
It increases
B.
It decreases
C.
It remains the same
D.
It fluctuates
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Solution
The vapor pressure of a solvent decreases when a non-volatile solute is added due to the solute particles occupying space at the surface.
Correct Answer: B — It decreases
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Q. What is the boiling point elevation of a solution containing 1 mol of NaCl in 1 kg of water? (K_b for water = 0.512 °C kg/mol)
A.
0.512 °C
B.
1.024 °C
C.
1.536 °C
D.
2.048 °C
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Solution
Boiling point elevation = i * K_b * m = 2 * 0.512 * 1 = 1.024 °C (i = 2 for NaCl)
Correct Answer: B — 1.024 °C
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Q. What is the boiling point of a solution containing 0.5 mol of KCl in 1 kg of water? (K_b for water = 0.512 °C kg/mol)
A.
1.024 °C
B.
0.512 °C
C.
1.536 °C
D.
2.048 °C
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Solution
Boiling point elevation = i * K_b * m = 2 * 0.512 * 0.5 = 0.512 °C; Boiling point = 100 + 0.512 = 100.512 °C
Correct Answer: C — 1.536 °C
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Q. What is the effect of a non-volatile solute on the boiling point of a solvent?
A.
It decreases the boiling point
B.
It increases the boiling point
C.
It has no effect
D.
It changes the boiling point unpredictably
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Solution
The addition of a non-volatile solute raises the boiling point of the solvent, a phenomenon known as boiling point elevation.
Correct Answer: B — It increases the boiling point
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Q. What is the expected osmotic pressure of a 0.5 M NaCl solution at 25 °C?
A.
12.3 atm
B.
24.6 atm
C.
6.1 atm
D.
3.1 atm
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Solution
Osmotic pressure (π) can be calculated using the formula π = iCRT. For NaCl, i = 2, C = 0.5 M, R = 0.0821 L·atm/(K·mol), and T = 298 K, resulting in approximately 24.6 atm.
Correct Answer: B — 24.6 atm
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Q. What is the formula for calculating boiling point elevation?
A.
ΔT_b = K_b * m
B.
ΔT_b = K_f * m
C.
ΔT_b = i * K_b * m
D.
ΔT_b = i * K_f * m
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Solution
The boiling point elevation is calculated using the formula ΔT_b = i * K_b * m, where i is the van 't Hoff factor.
Correct Answer: C — ΔT_b = i * K_b * m
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Q. What is the formula for calculating the depression of freezing point?
A.
ΔTf = Kf * m
B.
ΔTf = Kb * m
C.
ΔTf = R * T
D.
ΔTf = P * V
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Solution
The depression of freezing point is calculated using the formula ΔTf = Kf * m, where Kf is the freezing point depression constant and m is the molality of the solution.
Correct Answer: A — ΔTf = Kf * m
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Q. What is the freezing point depression constant (Kf) for water?
A.
1.86 °C kg/mol
B.
0.52 °C kg/mol
C.
2.00 °C kg/mol
D.
3.72 °C kg/mol
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Solution
The freezing point depression constant (Kf) for water is 1.86 °C kg/mol.
Correct Answer: A — 1.86 °C kg/mol
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Q. What is the freezing point depression of a solution directly proportional to?
A.
The molar mass of the solute
B.
The number of solute particles
C.
The volume of the solvent
D.
The temperature of the solvent
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Solution
Freezing point depression is directly proportional to the number of solute particles in the solution.
Correct Answer: B — The number of solute particles
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Q. What is the freezing point depression of a solution if 0.5 mol of a non-volatile solute is dissolved in 1 kg of water? (Kf for water = 1.86 °C kg/mol)
A.
0.93 °C
B.
1.86 °C
C.
3.72 °C
D.
0.5 °C
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Solution
Freezing point depression = Kf * molality = 1.86 * 0.5 = 0.93 °C.
Correct Answer: A — 0.93 °C
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Q. What is the freezing point of a solution containing 0.3 mol of glucose in 1 kg of water? (K_f for water = 1.86 °C kg/mol)
A.
-0.558 °C
B.
-0.558 K
C.
-1.86 °C
D.
-1.86 K
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Solution
Freezing point depression = K_f * m = 1.86 * 0.3 = 0.558 °C; Freezing point = 0 - 0.558 = -0.558 °C
Correct Answer: A — -0.558 °C
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Q. What is the osmotic pressure of a 0.2 M NaCl solution at 25 °C? (R = 0.0821 L atm/(K mol))
A.
4.92 atm
B.
2.46 atm
C.
1.23 atm
D.
0.61 atm
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Solution
Osmotic pressure = iCRT = 2 * 0.2 * 0.0821 * 298 = 4.92 atm (i = 2 for NaCl)
Correct Answer: A — 4.92 atm
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Q. What is the osmotic pressure of a solution containing 0.2 moles of solute in 1 liter of solution at 25 °C? (R = 0.0821 L·atm/(K·mol))
A.
4.92 atm
B.
1.64 atm
C.
0.82 atm
D.
2.46 atm
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Solution
Osmotic pressure = (n/V)RT = (0.2/1) * 0.0821 * 298 = 4.92 atm.
Correct Answer: A — 4.92 atm
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Q. What is the osmotic pressure of a solution directly proportional to?
A.
Temperature
B.
Concentration of solute
C.
Volume of solvent
D.
Both A and B
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Solution
Osmotic pressure is directly proportional to both the temperature and the concentration of solute.
Correct Answer: D — Both A and B
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Q. What is the primary factor affecting colligative properties?
A.
Nature of solute
B.
Concentration of solute
C.
Temperature
D.
Volume of solvent
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Solution
Colligative properties depend primarily on the concentration of solute particles in a solution.
Correct Answer: B — Concentration of solute
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Q. What is the primary reason for the elevation of boiling point in a solution?
A.
Increased molecular weight
B.
Decreased vapor pressure
C.
Increased solubility
D.
Decreased temperature
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Solution
The elevation of boiling point occurs because the presence of solute lowers the vapor pressure of the solvent, requiring a higher temperature to reach boiling.
Correct Answer: B — Decreased vapor pressure
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