Concentration Terms for Competitive Exam Success

Concentration Terms

Q. If 5 moles of KCl are dissolved in 3 kg of water, what is the molality of the solution?

A. 1.67 m
B. 2.5 m
C. 1.25 m
D. 0.5 m

Solution

Molality (m) = moles of solute / kg of solvent = 5 moles / 3 kg = 1.67 m.

Correct Answer: A — 1.67 m

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Q. If 50 g of KCl is dissolved in 250 mL of solution, what is the molarity? (Molar mass of KCl = 74.5 g/mol)

A. 2.68 M
B. 1.34 M
C. 0.67 M
D. 0.25 M

Solution

Moles of KCl = 50 g / 74.5 g/mol = 0.672 moles. Molarity = 0.672 moles / 0.25 L = 2.688 M.

Correct Answer: B — 1.34 M

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Q. If 50 grams of a solute is dissolved in 200 mL of solution, what is the mass/volume percent concentration?

A. 25%
B. 20%
C. 15%
D. 10%

Solution

Mass/volume percent = (mass of solute / volume of solution) * 100 = (50 g / 200 mL) * 100 = 25%.

Correct Answer: A — 25%

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Q. If 50 grams of NaCl is dissolved in 1 liter of solution, what is the molarity? (Molar mass of NaCl = 58.5 g/mol)

A. 0.85 M
B. 1.0 M
C. 1.5 M
D. 2.0 M

Solution

Moles of NaCl = 50 g / 58.5 g/mol = 0.854 moles. Molarity = 0.854 moles / 1 L = 0.854 M.

Correct Answer: B — 1.0 M

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Q. If 50 grams of NaOH is dissolved in 1 liter of solution, what is the molarity? (Molar mass of NaOH = 40 g/mol)

A. 1.25 M
B. 2.5 M
C. 0.5 M
D. 1 M

Solution

Moles of NaOH = 50 g / 40 g/mol = 1.25 moles. Molarity = 1.25 moles / 1 L = 1.25 M.

Correct Answer: B — 2.5 M

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Q. If 50 mL of a 2 M HCl solution is diluted to 200 mL, what is the new molarity?

A. 0.25 M
B. 0.5 M
C. 1 M
D. 2 M

Solution

Using the dilution formula M1V1 = M2V2, (2 M)(50 mL) = M2(200 mL) => M2 = 0.5 M.

Correct Answer: B — 0.5 M

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Q. What is the concentration in g/L of a solution containing 0.5 moles of KCl in 1.5 L of solution? (Molar mass of KCl = 74.5 g/mol)

A. 24.83 g/L
B. 49.67 g/L
C. 37.25 g/L
D. 50 g/L

Solution

Mass of KCl = 0.5 moles × 74.5 g/mol = 37.25 g. Concentration = 37.25 g / 1.5 L = 24.83 g/L.

Correct Answer: B — 49.67 g/L

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Q. What is the concentration in molality of a solution made by dissolving 5 moles of solute in 3 kg of solvent?

A. 1.67 m
B. 2 m
C. 1.5 m
D. 2.5 m

Solution

Molality (m) = moles of solute / kg of solvent = 5 moles / 3 kg = 1.67 m.

Correct Answer: B — 2 m

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Q. What is the concentration in molarity of a solution that contains 0.5 moles of solute in 1.5 liters of solution?

A. 0.33 M
B. 0.5 M
C. 0.75 M
D. 1 M

Solution

Molarity (M) = moles of solute / liters of solution = 0.5 moles / 1.5 L = 0.33 M.

Correct Answer: C — 0.75 M

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Q. What is the concentration in ppm of a solution containing 1 gram of solute in 1 liter of solution?

A. 1000 ppm
B. 100 ppm
C. 10 ppm
D. 1 ppm

Solution

PPM = (mass of solute in grams / volume of solution in liters) * 10^6 = (1 g / 1 L) * 10^6 = 1000 ppm.

Correct Answer: A — 1000 ppm

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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

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 mass percent of a solution containing 20 g of NaCl in 180 g of water?

A. 10%
B. 20%
C. 25%
D. 15%

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%

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%

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

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

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 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

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

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

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

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 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

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

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

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?

A. 4 N
B. 8 N
C. 2 N
D. 1 N

Solution

Normality (N) = equivalents of solute / liters of solution. H2SO4 has 2 equivalents, so 4 moles = 8 equivalents. N = 8 eq / 2 L = 4 N.

Correct Answer: B — 8 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

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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Q. What is the normality of a solution that is 1 M in H2SO4?

A. 1 N
B. 2 N
C. 0.5 N
D. 4 N

Solution

Normality (N) = Molarity (M) x number of equivalents. H2SO4 has 2 acidic protons, so 1 M x 2 = 2 N.

Correct Answer: B — 2 N

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Q. What is the percentage by mass of a solution containing 20 grams of solute in 180 grams of solution?

A. 10%
B. 20%
C. 15%
D. 25%

Solution

Percentage by mass = (mass of solute / mass of solution) * 100 = (20 g / (20 g + 180 g)) * 100 = 20%.

Correct Answer: B — 20%

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Q. What is the percentage by mass of a solution containing 5 g of NaCl in 95 g of water?

A. 5%
B. 10%
C. 15%
D. 20%

Solution

Percentage by mass = (mass of solute / total mass) × 100 = (5 g / (5 g + 95 g)) × 100 = 5%.

Correct Answer: A — 5%

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Q. What is the volume of 1 M NaOH solution required to obtain 0.5 moles of NaOH?

A. 0.5 L
B. 1 L
C. 2 L
D. 0.25 L

Solution

Using the formula M = moles/volume, Volume = moles/M = 0.5 moles / 1 M = 0.5 L.

Correct Answer: A — 0.5 L

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Q. What is the volume of a 2 M solution that contains 4 moles of solute?

A. 2 L
B. 4 L
C. 1 L
D. 0.5 L

Solution

Volume = moles of solute / molarity = 4 moles / 2 M = 2 L.

Correct Answer: B — 4 L

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Showing 31 to 60 of 69 (3 Pages)

Concentration Terms

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