Q. How many grams of KCl are needed to prepare 250 mL of a 0.5 M solution? (2023) 2023
A.
7.45 g
B.
12.5 g
C.
9.25 g
D.
5.25 g
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Solution
Molar mass of KCl = 74.55 g/mol. Grams = Molarity * Volume (L) * Molar mass = 0.5 M * 0.25 L * 74.55 g/mol = 9.31 g.
Correct Answer:
A
— 7.45 g
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Q. How many grams of KCl are needed to prepare 250 mL of a 2 M solution? (2023)
A.
37.25 g
B.
50 g
C.
75 g
D.
25 g
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Solution
Molar mass of KCl = 74.55 g/mol. Grams = moles × molar mass = 0.5 moles × 74.55 g/mol = 37.25 g.
Correct Answer:
A
— 37.25 g
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Q. How many grams of KCl are needed to prepare 250 mL of a 2 M solution? (Molar mass of KCl = 74.5 g/mol) (2023)
A.
37.25 g
B.
74.5 g
C.
18.625 g
D.
9.25 g
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Solution
Grams = moles × molar mass = (2 moles/L × 0.25 L) × 74.5 g/mol = 37.25 g.
Correct Answer:
A
— 37.25 g
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Q. How many liters of a 0.5 M NaOH solution are needed to obtain 1 mole of NaOH? (2023)
A.
1 L
B.
2 L
C.
0.5 L
D.
0.25 L
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Solution
Volume = moles / molarity = 1 mole / 0.5 M = 2 L.
Correct Answer:
A
— 1 L
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Q. How many liters of a 0.5 M solution contain 2 moles of solute? (2023)
A.
2 L
B.
4 L
C.
1 L
D.
0.5 L
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Solution
Volume = moles / molarity = 2 moles / 0.5 M = 4 L
Correct Answer:
B
— 4 L
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Q. How many liters of a 3 M solution can be made from 6 moles of solute? (2023)
A.
1 L
B.
2 L
C.
3 L
D.
4 L
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Solution
Volume = moles / molarity = 6 moles / 3 M = 2 L.
Correct Answer:
B
— 2 L
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Q. How many moles of solute are present in 500 mL of a 0.2 M solution? (2021) 2021
A.
0.1 moles
B.
0.5 moles
C.
0.2 moles
D.
0.25 moles
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Solution
Moles = Molarity * Volume (L) = 0.2 M * 0.5 L = 0.1 moles.
Correct Answer:
A
— 0.1 moles
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Q. If 10 g of glucose (C6H12O6) is dissolved in 200 g of water, what is the mass percentage of the solution? (2023)
A.
4.76%
B.
5.00%
C.
10.00%
D.
2.50%
Show solution
Solution
Mass percentage = (mass of solute / mass of solution) x 100. Mass of solution = 10 g + 200 g = 210 g. Mass percentage = (10 g / 210 g) x 100 = 4.76%.
Correct Answer:
A
— 4.76%
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Q. If 10 g of sugar (C12H22O11) is dissolved in 200 g of water, what is the mass percent of sugar in the solution? (Molar mass of sugar = 342 g/mol) (2023)
A.
4.76%
B.
5.00%
C.
10.00%
D.
2.50%
Show solution
Solution
Mass percent = (mass of solute / total mass of solution) x 100. Total mass = 10 g + 200 g = 210 g. Mass percent = (10 g / 210 g) x 100 = 4.76%.
Correct Answer:
A
— 4.76%
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Q. If 10 g of sugar (C12H22O11) is dissolved in 200 g of water, what is the mass percent of the sugar solution? (2023)
A.
4.76%
B.
5.00%
C.
10.00%
D.
2.50%
Show solution
Solution
Mass percent = (mass of solute / mass of solution) x 100. Mass of solution = 10 g + 200 g = 210 g. Mass percent = (10 g / 210 g) x 100 = 4.76%.
Correct Answer:
A
— 4.76%
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Q. If 10 g of sugar (C12H22O11) is dissolved in 200 g of water, what is the mass percent of the solution? (Molar mass of sugar = 342 g/mol) (2023)
A.
4.76%
B.
5.00%
C.
10.00%
D.
2.50%
Show solution
Solution
Mass percent = (mass of solute / mass of solution) x 100. Mass of solution = 10 g + 200 g = 210 g. Mass percent = (10 g / 210 g) x 100 = 4.76%.
Correct Answer:
B
— 5.00%
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Q. If 10 g of sugar (C12H22O11) is dissolved in 200 g of water, what is the mass percent of the solution? (2023)
A.
4.76%
B.
5.00%
C.
10.00%
D.
20.00%
Show solution
Solution
Mass percent = (mass of solute / mass of solution) × 100. Mass of solution = 10 g + 200 g = 210 g. Mass percent = (10 g / 210 g) × 100 = 4.76%.
Correct Answer:
B
— 5.00%
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Q. If 10 g of sugar (C12H22O11) is dissolved in 200 g of water, what is the mass percentage of the solution? (Molar mass of sugar = 342 g/mol) (2023)
A.
4.76%
B.
5.00%
C.
10.00%
D.
2.50%
Show solution
Solution
Mass percentage = (mass of solute / mass of solution) x 100. Mass of solution = 10 g + 200 g = 210 g. Mass percentage = (10 g / 210 g) x 100 = 4.76%.
Correct Answer:
B
— 5.00%
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Q. If 10 g of sugar (C12H22O11) is dissolved in 200 g of water, what is the mass percent of the sugar solution? (Molar mass of sugar = 342 g/mol) (2023)
A.
4.76%
B.
5.00%
C.
10.00%
D.
2.50%
Show solution
Solution
Mass percent = (mass of solute / mass of solution) × 100. Mass of solution = 10 g + 200 g = 210 g. Mass percent = (10 g / 210 g) × 100 = 4.76%.
Correct Answer:
A
— 4.76%
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Q. If 10 g of sugar (C12H22O11) is dissolved in 200 g of water, what is the mass percentage of the sugar solution? (Molar mass of sugar = 342 g/mol) (2023)
A.
4.76%
B.
5.00%
C.
10.00%
D.
2.50%
Show solution
Solution
Mass percentage = (mass of solute / mass of solution) x 100. Mass of solution = 10 g + 200 g = 210 g. Mass percentage = (10 g / 210 g) x 100 = 4.76%.
Correct Answer:
A
— 4.76%
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Q. If 10 g of sugar is dissolved in 100 g of water, what is the mass percent of sugar in the solution? (2023)
A.
10%
B.
5%
C.
20%
D.
15%
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Solution
Mass percent = (mass of solute / total mass of solution) x 100. Total mass = 10 g + 100 g = 110 g. Mass percent = (10 g / 110 g) x 100 = 9.09%, approximately 10%.
Correct Answer:
A
— 10%
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Q. If 10 g of sugar is dissolved in 100 g of water, what is the mass percentage of sugar in the solution? (2023)
A.
10%
B.
5%
C.
20%
D.
15%
Show solution
Solution
Mass percentage = (mass of solute / mass of solution) × 100. Mass of solution = 10 g + 100 g = 110 g. Mass percentage = (10 g / 110 g) × 100 = 9.09%, approximately 10%.
Correct Answer:
A
— 10%
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Q. If 10 grams of NaCl is dissolved in 500 mL of water, what is the concentration in g/L? (2020)
A.
20 g/L
B.
10 g/L
C.
5 g/L
D.
15 g/L
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Solution
Concentration = (10 g / 0.5 L) = 20 g/L.
Correct Answer:
A
— 20 g/L
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Q. If 10 grams of NaCl is dissolved in 500 mL of water, what is the mass/volume percent concentration? (2020) 2020
A.
2%
B.
5%
C.
10%
D.
20%
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Solution
Mass/volume % = (mass of solute / volume of solution) * 100 = (10 g / 500 mL) * 100 = 2%.
Correct Answer:
B
— 5%
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Q. If 15 g of sugar is dissolved in 300 mL of water, what is the concentration in g/mL? (2020)
A.
0.05 g/mL
B.
0.1 g/mL
C.
0.15 g/mL
D.
0.2 g/mL
Show solution
Solution
Concentration = mass / volume = 15 g / 300 mL = 0.05 g/mL.
Correct Answer:
B
— 0.1 g/mL
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Q. If 20 grams of glucose (C6H12O6) is dissolved in 1 liter of water, what is the molarity of the solution? (Molar mass of glucose = 180 g/mol) (2020) 2020
A.
0.11 M
B.
0.5 M
C.
0.2 M
D.
0.33 M
Show solution
Solution
Moles of glucose = 20 g / 180 g/mol = 0.111 M. Molarity = 0.111 moles / 1 L = 0.11 M.
Correct Answer:
A
— 0.11 M
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Q. If 3 moles of glucose are dissolved in 1 liter of solution, what is the molality of the solution? (2023)
A.
3 m
B.
1 m
C.
2 m
D.
0.5 m
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Solution
Molality (m) = moles of solute / kg of solvent; assuming 1 L of water = 1 kg, m = 3 moles / 1 kg = 3 m
Correct Answer:
A
— 3 m
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Q. If 50 mL of a 3 M solution is diluted to 200 mL, what is the new molarity? (2020)
A.
0.75 M
B.
1.5 M
C.
2 M
D.
3 M
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Solution
Using M1V1 = M2V2, (3 M)(50 mL) = M2(200 mL) => M2 = (3 M × 50 mL) / 200 mL = 0.75 M.
Correct Answer:
A
— 0.75 M
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Q. What is the concentration in mol/L of a solution made by dissolving 58.5 grams of NaCl in 1 liter of water? (Molar mass of NaCl = 58.5 g/mol) (2023) 2023
A.
1 M
B.
0.5 M
C.
2 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. What is the effect of adding a non-volatile solute to a solvent? (2023)
A.
Increases vapor pressure
B.
Decreases boiling point
C.
Increases boiling point
D.
Decreases freezing point
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Solution
Adding a non-volatile solute to a solvent increases the boiling point of the solution due to the colligative property of boiling point elevation.
Correct Answer:
C
— Increases boiling point
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Q. What is the effect of increasing the temperature on the solubility of most solid solutes in water? (2023)
A.
Increases solubility
B.
Decreases solubility
C.
No effect
D.
Depends on the solute
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Solution
For most solid solutes, increasing the temperature increases their solubility in water.
Correct Answer:
A
— Increases solubility
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Q. What is the effect of temperature on the solubility of most solid solutes in water? (2023)
A.
Increases with temperature
B.
Decreases with temperature
C.
Remains constant
D.
Varies unpredictably
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Solution
For most solid solutes, solubility increases with an increase in temperature.
Correct Answer:
A
— Increases with temperature
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Q. What is the final volume when 100 mL of a 3 M solution is diluted to a concentration of 1 M? (2022) 2022
A.
300 mL
B.
200 mL
C.
100 mL
D.
150 mL
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Solution
Using the dilution formula C1V1 = C2V2, we have 3 M * 100 mL = 1 M * V2. Thus, V2 = 300 mL.
Correct Answer:
A
— 300 mL
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Q. What is the freezing point depression of a solution containing 1 mol of NaCl in 1 kg of water? (Kf for water = 1.86 °C kg/mol) (2023)
A.
1.86 °C
B.
3.72 °C
C.
2.00 °C
D.
0.93 °C
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Solution
Freezing point depression (ΔTf) = i * Kf * m. For NaCl, i = 2 (dissociates into Na+ and Cl-). m = 1 mol/kg. ΔTf = 2 * 1.86 °C kg/mol * 1 = 3.72 °C.
Correct Answer:
B
— 3.72 °C
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Q. What is the mass of NaOH required to prepare 0.5 L of a 1 M solution? (2023)
A.
20 g
B.
40 g
C.
10 g
D.
30 g
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Solution
Molar mass of NaOH = 40 g/mol; grams = moles × molar mass = (1 mole/L × 0.5 L) × 40 g/mol = 20 g
Correct Answer:
B
— 40 g
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