Solutions
Q. A solution contains 20% (w/w) of glucose. If the total mass of the solution is 200 g, what is the mass of glucose in the solution?
A.
20 g
B.
40 g
C.
60 g
D.
80 g
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Solution
Mass of glucose = (20/100) x 200 g = 40 g.
Correct Answer: B — 40 g
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Q. A solution has a concentration of 0.1 M NaCl. How many grams of NaCl are present in 1 liter of this solution? (Molar mass of NaCl = 58.5 g/mol)
A.
5.85 g
B.
58.5 g
C.
0.1 g
D.
0.585 g
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Solution
Mass = moles x molar mass = 0.1 moles x 58.5 g/mol = 5.85 g.
Correct Answer: B — 58.5 g
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Q. A solution has a concentration of 0.2 M. How many moles of solute are present in 1.5 L of this solution?
A.
0.3 moles
B.
0.5 moles
C.
0.2 moles
D.
0.15 moles
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Solution
Moles of solute = Molarity × Volume = 0.2 M × 1.5 L = 0.3 moles.
Correct Answer: B — 0.5 moles
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Q. A solution has a density of 1.2 g/mL and contains 10% (w/v) NaOH. What is the mass of NaOH in 1 L of this solution?
A.
100 g
B.
120 g
C.
80 g
D.
60 g
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Solution
Mass of NaOH = (10/100) x 1000 mL = 100 g.
Correct Answer: B — 120 g
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Q. A solution has a density of 1.2 g/mL and contains 30 g of solute. What is the molarity if the molar mass of the solute is 60 g/mol?
A.
0.5 M
B.
1 M
C.
2 M
D.
1.5 M
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Solution
Volume of solution = mass / density = 30 g / 1.2 g/mL = 25 mL = 0.025 L. Moles of solute = 30 g / 60 g/mol = 0.5 moles. Molarity = 0.5 moles / 0.025 L = 20 M.
Correct Answer: B — 1 M
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Q. A solution is prepared by dissolving 50 g of glucose (C6H12O6) in 250 g of water. What is the mass percent of glucose in the solution? (Molar mass of glucose = 180 g/mol)
A.
20%
B.
15%
C.
25%
D.
10%
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Solution
Mass percent = (mass of solute / (mass of solute + mass of solvent)) × 100 = (50 g / (50 g + 250 g)) × 100 = 20%.
Correct Answer: A — 20%
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Q. A solution is prepared by dissolving 58.5 g of NaCl in 1 L of water. What is the molarity of the solution? (Molar mass of NaCl = 58.5 g/mol)
A.
1 M
B.
2 M
C.
0.5 M
D.
0.1 M
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Solution
Moles of NaCl = 58.5 g / 58.5 g/mol = 1 mole. Molarity = moles of solute / liters of solution = 1 mole / 1 L = 1 M.
Correct Answer: A — 1 M
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Q. A solution is prepared by dissolving 58.5 g of NaCl in 1 L of water. What is the concentration in terms of molarity? (Molar mass of NaCl = 58.5 g/mol)
A.
1 M
B.
2 M
C.
0.5 M
D.
0.25 M
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Solution
Moles of NaCl = 58.5 g / 58.5 g/mol = 1 mole. Molarity = 1 mole / 1 L = 1 M.
Correct Answer: A — 1 M
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Q. A solution is prepared by dissolving 58.5 g of NaCl in enough water to make 1 L of solution. What is the molarity of the solution? (Molar mass of NaCl = 58.5 g/mol)
A.
1 M
B.
2 M
C.
0.5 M
D.
0.1 M
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Solution
Moles of NaCl = 58.5 g / 58.5 g/mol = 1 mole. Molarity = 1 mole / 1 L = 1 M.
Correct Answer: A — 1 M
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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. For a solution to obey Raoult's Law, the interactions between solute and solvent must be:
A.
Stronger than those in the pure components.
B.
Weaker than those in the pure components.
C.
Similar to those in the pure components.
D.
Non-existent.
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Solution
For a solution to obey Raoult's Law, the interactions between solute and solvent must be similar to those in the pure components.
Correct Answer: C — Similar to those in the pure components.
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Q. For an ideal solution, if the mole fraction of the solvent is 0.75, what is the vapor pressure of the solution if the vapor pressure of the pure solvent is 100 mmHg?
A.
75 mmHg
B.
100 mmHg
C.
25 mmHg
D.
50 mmHg
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Solution
According to Raoult's Law, the vapor pressure of the solution is 0.75 * 100 mmHg = 75 mmHg.
Correct Answer: A — 75 mmHg
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Q. For the equation x^2 + 2x + k = 0 to have one root equal to 1, what is the value of k?
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Solution
Substituting x = 1 into the equation gives 1^2 + 2*1 + k = 0 => 1 + 2 + k = 0 => k = -3.
Correct Answer: B — 1
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Q. For the equation x^2 + kx + 9 = 0 to have real roots, what must be true about k?
A.
k < 6
B.
k > 6
C.
k < 0
D.
k > 0
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Solution
The discriminant must be non-negative: k^2 - 4*1*9 >= 0 => k^2 >= 36 => k <= -6 or k >= 6.
Correct Answer: A — k < 6
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Q. For which value of k does the equation x^2 + kx + 16 = 0 have no real roots?
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Solution
The discriminant must be less than zero: k^2 - 4*1*16 < 0 => k^2 < 64 => k < 8 and k > -8.
Correct Answer: A — -8
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Q. How many grams of solute are needed to prepare 0.5 L of a 1 M NaCl solution? (Molar mass of NaCl = 58.5 g/mol)
A.
29.25 g
B.
58.5 g
C.
14.625 g
D.
0.5 g
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Solution
Mass of solute = moles × molar mass = 1 mol × 58.5 g/mol = 58.5 g. For 0.5 L, it is 0.5 mol × 58.5 g/mol = 29.25 g.
Correct Answer: A — 29.25 g
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Q. How many grams of solute are needed to prepare 2 L of a 1 M solution? (Molar mass = 58.5 g/mol)
A.
58.5 g
B.
117 g
C.
29.25 g
D.
145 g
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Solution
Grams of solute = moles x molar mass = 2 moles x 58.5 g/mol = 117 g.
Correct Answer: B — 117 g
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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 L of a 2 M solution is diluted to 3 L, what is the new molarity of the solution?
A.
0.67 M
B.
1 M
C.
1.5 M
D.
2 M
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Solution
Using the dilution formula M1V1 = M2V2, we have 2 M * 1 L = M2 * 3 L, thus M2 = 2/3 = 0.67 M.
Correct Answer: A — 0.67 M
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Q. If 1 L of a 3 M solution is diluted to 2 L, what is the new molarity?
A.
1.5 M
B.
3 M
C.
6 M
D.
0.5 M
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Solution
Using the dilution formula M1V1 = M2V2, we have 3 M × 1 L = M2 × 2 L. Thus, M2 = 1.5 M.
Correct Answer: A — 1.5 M
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Q. If 1 liter of a 2 M solution is diluted to 3 liters, what is the new molarity?
A.
0.67 M
B.
1 M
C.
1.5 M
D.
2 M
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Solution
Using the dilution formula M1V1 = M2V2, (2 M)(1 L) = M2(3 L) => M2 = 2/3 = 0.67 M.
Correct Answer: A — 0.67 M
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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 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 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-volatile solute is dissolved in 1 kg of water, what is the expected change in boiling point? (Kb for water = 0.512 °C kg/mol)
A.
0.512 °C
B.
1.024 °C
C.
2.048 °C
D.
0.256 °C
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Solution
Boiling point elevation = i * Kb * m = 1 * 0.512 * 1 = 0.512 °C.
Correct Answer: B — 1.024 °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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