Q: A student has a stock solution of 5.0 M KCl. What is the final volume required to dilute 100 mL of this solution to a concentration of 0.54 M KCl?

$$ \frac{M_1V_1}{M_2$} = \frac{5.0 \times 100}{0.54} = 270 \text{ mL} $$

Q: If 880 mL of water is added to 125.0 mL of a 0.770 M HBrO$$_3$$ solution, what is the resulting molarity?

$$ M_2$ = \frac{M_1V_1}{V_2$} = \frac{0.770 \times 125.0}{1005.0} = 0.0957 \text{ M} $$

Q: A student prepares a stock solution by dissolving 25.0 g of NaOH in enough water to make 150.0 mL solution. They take 20.0 mL of this stock and dilute it to 250.0 mL. Then, 75.0 mL of this diluted solution is further diluted to 500.0 mL. What is the final concentration of NaOH in mol/L? (MM NaOH = 40.00 g/mol)

$$ \text{Final concentration} = \frac{25.0}{40.00} \times \frac{20.0}{150.0} \times \frac{75.0}{250.0} \times \frac{1}{500.0} = 0.00112 \text{ M} $$

Q: What volume (in mL) of 5.2 M HBr must be used to prepare 3.5 L of 2.7 M HBr?

$$ V_1$ = \frac{M_2V_2}{M_1$} = \frac{2.7 \times 3500}{5.2} = 1817.3 \text{ mL} $$

Q: If you have 250 mL of a 0.80 M solution and you dilute it to 1.00 L, what is the final concentration?

$$ M_2$ = \frac{M_1V_1}{V_2$} = \frac{0.80 \times 250}{1000} = 0.20 \text{ M} $$

Q: A nurse needs to prepare 500 mL of a 0.1 M NaCl solution from a 2.0 M stock. What volume of stock solution is required?

$$ V_1$ = \frac{M_2V_2}{M_1$} = \frac{0.1 \times 500}{2.0} = 25 \text{ mL} $$

Q: A solution is prepared by diluting 50 mL of 3.0 M H$$_2SO$_4 to a $$final volume of 600 mL. What is the final concentration?

$$ M_2$ = \frac{M_1V_1}{V_2$} = \frac{3.0 \times 50}{600} = 0.25 \text{ M} $$

Question 1

Which of the following best describes the process of dilution in a laboratory setting?

Accepted Answer

Adding more solvent to a solution to decrease its concentration.

Question 2

A student has a stock solution of 5.0 M KCl. What is the final volume required to dilute 100 mL of this solution to a concentration of 0.54 M KCl?

Accepted Answer

$$ \frac{M_1V_1}{M_2$} = \frac{5.0 	imes 100}{0.54} = 270 	ext{ mL} $$

Question 3

If 880 mL of water is added to 125.0 mL of a 0.770 M HBrO$$_3$$ solution, what is the resulting molarity?

Accepted Answer

$$ M_2$ = \frac{M_1V_1}{V_2$} = \frac{0.770 	imes 125.0}{1005.0} = 0.0957 	ext{ M} $$

Question 4

A student prepares a stock solution by dissolving 25.0 g of NaOH in enough water to make 150.0 mL solution. They take 20.0 mL of this stock and dilute it to 250.0 mL. Then, 75.0 mL of this diluted solution is further diluted to 500.0 mL. What is the final concentration of NaOH in mol/L? (MM NaOH = 40.00 g/mol)

Accepted Answer

$$ 	ext{Final concentration} = \frac{25.0}{40.00} 	imes \frac{20.0}{150.0} 	imes \frac{75.0}{250.0} 	imes \frac{1}{500.0} = 0.00112 	ext{ M} $$

Question 5

Which solution is the most concentrated if each sphere represents a mole of solute: Solution A (5 spheres/1 L), Solution B (3 spheres/1 L), Solution C (8 spheres/2 L)?

Accepted Answer

Solution C

Question 6

What volume (in mL) of 5.2 M HBr must be used to prepare 3.5 L of 2.7 M HBr?

Accepted Answer

$$ V_1$ = \frac{M_2V_2}{M_1$} = \frac{2.7 	imes 3500}{5.2} = 1817.3 	ext{ mL} $$

Question 7

Which of the following statements about a stock solution is correct?

Accepted Answer

A stock solution is a concentrated solution used to prepare more dilute solutions.

Question 8

If you have 250 mL of a 0.80 M solution and you dilute it to 1.00 L, what is the final concentration?

Accepted Answer

$$ M_2$ = \frac{M_1V_1}{V_2$} = \frac{0.80 	imes 250}{1000} = 0.20 	ext{ M} $$

Question 9

A nurse needs to prepare 500 mL of a 0.1 M NaCl solution from a 2.0 M stock. What volume of stock solution is required?

Accepted Answer

$$ V_1$ = \frac{M_2V_2}{M_1$} = \frac{0.1 	imes 500}{2.0} = 25 	ext{ mL} $$

Question 10

Which equation is used to calculate the final concentration after dilution?

Accepted Answer

$$ M_1V_1 = M_2V_2 $$

Question 11

A solution is prepared by diluting 50 mL of 3.0 M H$$_2SO$_4 to a $$final volume of 600 mL. What is the final concentration?

Accepted Answer

$$ M_2$ = \frac{M_1V_1}{V_2$} = \frac{3.0 	imes 50}{600} = 0.25 	ext{ M} $$

Question 12

Which of the following is a correct real-life application of dilution calculations?

Accepted Answer

Determining the amount of water needed to dilute a concentrated medication for intravenous use.

Question 13

If you have 100 mL of a 1.5 M solution and you want to dilute it to 0.5 M, what will be the final volume?

Accepted Answer

$$ V_2$ = \frac{M_1V_1}{M_2$} = \frac{1.5 	imes 100}{0.5} = 300 	ext{ mL} $$

Question 14

Which of the following would NOT affect the concentration of a solution during dilution?

Accepted Answer

Changing the container without adding or removing solvent or solute.

Question 15

A chemist needs to prepare 250 mL of 0.25 M solution from a 1.0 M stock. How much stock solution should be used?

Accepted Answer

$$ V_1$ = \frac{M_2V_2}{M_1$} = \frac{0.25 	imes 250}{1.0} = 62.5 	ext{ mL} $$

Question 16

Which of the following is the correct order of solutions from least to most concentrated, given: Solution A (2 moles/1 L), Solution B (4 moles/2 L), Solution C (3 moles/0.5 L)?

Accepted Answer

A, B, C

Dilution and Solution Concentration in GOB Chemistry

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