Q. What is the effect of increasing temperature on the vapor pressure of a solvent according to Raoult's Law?
A.
It decreases the vapor pressure.
B.
It increases the vapor pressure.
C.
It has no effect.
D.
It depends on the solute.
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Solution
Increasing temperature generally increases the vapor pressure of a solvent, as more molecules have enough energy to escape into the vapor phase.
Correct Answer: B — It increases the vapor pressure.
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Q. What is the effect of increasing the concentration of a solute on the freezing point of a solution?
A.
Freezing point increases
B.
Freezing point decreases
C.
No effect
D.
Depends on the solute
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Solution
Increasing the concentration of a solute lowers the freezing point of a solution.
Correct Answer: B — Freezing point decreases
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Q. What is the effect of temperature on Raoult's Law?
A.
It has no effect.
B.
It increases vapor pressure.
C.
It decreases vapor pressure.
D.
It only affects the solute.
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Solution
Temperature affects the vapor pressure of both the solvent and the solution, generally increasing vapor pressure with temperature.
Correct Answer: B — It increases vapor pressure.
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Q. What is the effect of temperature on the vapor pressure of a solvent according to Raoult's Law?
A.
Vapor pressure decreases with temperature
B.
Vapor pressure increases with temperature
C.
Vapor pressure remains constant with temperature
D.
Vapor pressure is independent of the solvent
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Solution
According to Raoult's Law, the vapor pressure of a solvent increases with an increase in temperature.
Correct Answer: B — Vapor pressure increases with temperature
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Q. What is the equivalent weight of H2SO4 if its molar mass is 98 g/mol?
A.
49 g
B.
98 g
C.
196 g
D.
24.5 g
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Solution
Equivalent weight = molar mass / number of equivalents = 98 g/mol / 2 = 49 g.
Correct Answer: A — 49 g
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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 containing 2 moles of KCl in 1 kg of water?
A.
-3.72 °C
B.
-1.86 °C
C.
-2.52 °C
D.
-4.0 °C
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Solution
Freezing point depression = i * Kf * m = 3 * 1.86 * 2 = 11.16 °C, so the freezing point is lowered by 3.72 °C.
Correct Answer: A — -3.72 °C
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Q. What is the freezing point depression of a solution containing 2 moles of KCl in 1 kg of water? (Kf for water = 1.86 °C kg/mol)
A.
3.72 °C
B.
1.86 °C
C.
2.0 °C
D.
5.58 °C
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Solution
Freezing point depression = i * Kf * m = 3 * 1.86 * 2 = 11.16 °C.
Correct Answer: A — 3.72 °C
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Q. What is the freezing point depression of a solution containing 2 moles of NaCl in 1 kg of water? (Kf for water = 1.86 °C kg/mol)
A.
3.72 °C
B.
1.86 °C
C.
2.72 °C
D.
5.72 °C
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Solution
Freezing point depression = i * Kf * m = 3 * 1.86 * 2 = 11.16 °C.
Correct Answer: A — 3.72 °C
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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 mass percent of a solution containing 20 g of NaCl in 180 g of water?
A.
10%
B.
20%
C.
25%
D.
15%
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Solution
Mass percent = (mass of solute / total mass) x 100 = (20 g / (20 g + 180 g)) x 100 = 10%.
Correct Answer: A — 10%
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Q. What is the mass percent of a solution containing 20 g of solute in 180 g of solution?
A.
10%
B.
20%
C.
25%
D.
15%
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Solution
Mass percent = (mass of solute / mass of solution) x 100 = (20 g / 200 g) x 100 = 10%.
Correct Answer: B — 20%
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Q. What is the mass percent of a solution containing 20 g of solute in 200 g of solution?
A.
10%
B.
20%
C.
5%
D.
15%
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Solution
Mass percent = (mass of solute / mass of solution) × 100 = (20 g / 200 g) × 100 = 10%.
Correct Answer: B — 20%
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Q. What is the molality of a solution containing 3 moles of KCl dissolved in 1 kg of water?
A.
3 m
B.
1.5 m
C.
2 m
D.
4 m
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Solution
Molality (m) = moles of solute / kg of solvent = 3 moles / 1 kg = 3 m.
Correct Answer: A — 3 m
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Q. What is the molality of a solution prepared by dissolving 3 moles of KCl in 1 kg of water?
A.
3 m
B.
1.5 m
C.
2 m
D.
4 m
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Solution
Molality (m) = moles of solute / kg of solvent = 3 moles / 1 kg = 3 m.
Correct Answer: A — 3 m
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Q. What is the molality of a solution prepared by dissolving 5 moles of NaCl in 2 kg of water?
A.
2.5 mol/kg
B.
5 mol/kg
C.
1.5 mol/kg
D.
3 mol/kg
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Solution
Molality (m) = moles of solute / mass of solvent (kg) = 5 moles / 2 kg = 2.5 mol/kg.
Correct Answer: A — 2.5 mol/kg
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Q. What is the molarity of a solution if 10 g of glucose (C6H12O6) is dissolved in 250 mL of water? (Molar mass = 180 g/mol)
A.
0.22 M
B.
0.5 M
C.
0.75 M
D.
1 M
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Solution
Moles of glucose = 10 g / 180 g/mol = 0.0556 moles. Molarity = 0.0556 moles / 0.25 L = 0.222 M.
Correct Answer: A — 0.22 M
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Q. What is the molarity of a solution if 10 grams of CaCl2 is dissolved in 250 mL of solution? (Molar mass of CaCl2 = 110 g/mol)
A.
0.25 M
B.
0.5 M
C.
1 M
D.
2 M
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Solution
Moles of CaCl2 = 10 g / 110 g/mol = 0.0909 moles. Molarity = 0.0909 moles / 0.25 L = 0.3636 M.
Correct Answer: B — 0.5 M
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Q. What is the molarity of a solution prepared by dissolving 5 moles of NaCl in 2 liters of water?
A.
2.5 M
B.
5 M
C.
10 M
D.
1 M
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Solution
Molarity (M) = moles of solute / liters of solution = 5 moles / 2 L = 2.5 M.
Correct Answer: A — 2.5 M
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Q. What is the mole fraction of solute in a solution containing 2 moles of solute and 8 moles of solvent?
A.
0.2
B.
0.25
C.
0.5
D.
0.1
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Solution
Mole fraction of solute = moles of solute / (moles of solute + moles of solvent) = 2 / (2 + 8) = 2 / 10 = 0.2.
Correct Answer: B — 0.25
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Q. What is the mole fraction of solute in a solution containing 3 moles of solute and 7 moles of solvent?
A.
0.3
B.
0.7
C.
0.5
D.
0.2
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Solution
Mole fraction of solute = moles of solute / total moles = 3 / (3 + 7) = 0.3.
Correct Answer: A — 0.3
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Q. What is the normality of a solution containing 1 mole of H2SO4 in 1 liter of solution?
A.
1 N
B.
2 N
C.
0.5 N
D.
4 N
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Solution
Normality (N) = equivalents of solute / liters of solution. H2SO4 provides 2 equivalents, so N = 2 moles / 1 L = 2 N.
Correct Answer: B — 2 N
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Q. What is the normality of a solution containing 2 moles of H2SO4 in 1 liter of solution?
A.
2 N
B.
4 N
C.
1 N
D.
0.5 N
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Solution
Normality (N) = equivalents of solute / liters of solution. H2SO4 provides 2 equivalents, so 2 moles × 2 = 4 N.
Correct Answer: B — 4 N
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Q. What is the normality of a solution containing 3 moles of H2SO4 in 2 liters of solution? (H2SO4 is a diprotic acid)
A.
3 N
B.
6 N
C.
1.5 N
D.
1 N
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Solution
Normality (N) = equivalents of solute / liters of solution. H2SO4 has 2 equivalents per mole, so 3 moles = 6 equivalents. Normality = 6 equivalents / 2 L = 3 N.
Correct Answer: B — 6 N
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Q. What is the normality of a solution containing 4 moles of H2SO4 in 2 liters of solution? (H2SO4 is a diprotic acid)
A.
4 N
B.
2 N
C.
8 N
D.
1 N
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Solution
Normality (N) = number of equivalents / liters of solution. H2SO4 has 2 equivalents per mole, so 4 moles = 8 equivalents. Normality = 8 equivalents / 2 L = 4 N.
Correct Answer: C — 8 N
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