Q: A laboratory technician needs to prepare 500.0 mL of a 0.250 M NaCl solution. What mass of NaCl (molar mass = 58.44 g/mol) should be weighed out?

$$0.250 \times 0.500 \times 58.44 = 7.305\ \text{g}$$

Q: A solution is made by dissolving 9.60 g of glucose (molar mass = 180.2 g/mol) in 250.0 g of water. What is the molality of the solution?

$$\frac{9.60}{180.2} \div 0.250 = 0.213\ \text{m}$$

Q: A 23.40 g sample of naphthalene (C$$_{10}H$_8) is $$dissolved in 99.60 g benzene (C$$_6H$_6). $$What is the mass percent of naphthalene in the solution?

$$\frac{23.40}{23.40 + 99.60} \times 100 = 19.02\%$$

Q: A 30.0% by mass solution of hydrogen peroxide (H$$_2O$_2$$, molar mass = 34.02 g/mol) has a density of 1.135 g/mL. What is the molality of H$$_2O$_2 in $$this solution?

$$\frac{8.82}{0.700} = 12.6\ \text{m}$$

Q: For the same 30.0% by mass H$$_2O$_2$$ solution (density = 1.135 g/mL), what is the molarity of H$$_2O$_2$$?

$$\frac{8.82}{0.881} = 10.0\ \text{M}$$

Q: Which of the following is the correct formula for parts per million (ppm) in a solution?

$$\text{ppm} = \frac{\text{mass solute (g)}}{\text{mass solution (g)}} \times 10^6$

Question 1

Which of the following statements correctly distinguishes between molarity and molality?

Accepted Answer

Molarity uses the volume of solution in the denominator, while molality uses the mass of solvent.

Question 2

A laboratory technician needs to prepare 500.0 mL of a 0.250 M NaCl solution. What mass of NaCl (molar mass = 58.44 g/mol) should be weighed out?

Accepted Answer

$$0.250 	imes 0.500 	imes 58.44 = 7.305\ 	ext{g}$$

Question 3

Which of the following best describes how to prepare 250.0 mL of 0.290 M sucrose solution (molar mass = 342.3 g/mol)?

Accepted Answer

Dissolve 24.8 g of sucrose in about 150 mL of water, transfer to a 250 mL volumetric flask, dilute to the mark, and mix.

Question 4

A solution is made by dissolving 9.60 g of glucose (molar mass = 180.2 g/mol) in 250.0 g of water. What is the molality of the solution?

Accepted Answer

$$\frac{9.60}{180.2} \div 0.250 = 0.213\ 	ext{m}$$

Question 5

Calculate the mole fraction of glucose in a solution containing 8.60 g glucose (molar mass = 180.2 g/mol) dissolved in 26.5 g water (molar mass = 18.02 g/mol).

Accepted Answer

$$\frac{0.0477}{0.0477 + 1.4705} = 0.0314$$

Question 6

A 23.40 g sample of naphthalene (C$$_{10}H$_8) is $$dissolved in 99.60 g benzene (C$$_6H$_6). $$What is the mass percent of naphthalene in the solution?

Accepted Answer

$$\frac{23.40}{23.40 + 99.60} 	imes 100 = 19.02\%$$

Question 7

A 30.0% by mass solution of hydrogen peroxide (H$$_2O$_2$$, molar mass = 34.02 g/mol) has a density of 1.135 g/mL. What is the molality of H$$_2O$_2 in $$this solution?

Accepted Answer

$$\frac{8.82}{0.700} = 12.6\ 	ext{m}$$

Question 8

For the same 30.0% by mass H$$_2O$_2$$ solution (density = 1.135 g/mL), what is the molarity of H$$_2O$_2$$?

Accepted Answer

$$\frac{8.82}{0.881} = 10.0\ 	ext{M}$$

Question 9

A 10% (m/v) NaCl solution means:

Accepted Answer

10 g of NaCl dissolved in enough water to make 100 mL of solution.

Question 10

Which of the following is the correct formula for parts per million (ppm) in a solution?

Accepted Answer

$$	ext{ppm} = \frac{	ext{mass solute (g)}}{	ext{mass solution (g)}} 	imes 10^6$

Question 11

A 10.40 mL sample of lake water contains 20.60 μg of copper. What is the concentration of copper in ppm? (Assume density of water is 1.00 g/mL)

Accepted Answer

$$\frac{20.60 	imes 10$$^{-6}$$}{10.40 	imes 10$$^{-3}$$} 	imes 10^6$ = 1.98\ 	ext{ppm}$$

Question 12

A 15.68 mL sample of drinking water contains 0.01800 μg of lead. What is the concentration of lead in ppb? (Assume density of water is 1.00 g/mL)

Accepted Answer

$$\frac{0.01800 	imes 10$$^{-6}$$}{15.68 	imes 10$$^{-3}$$} 	imes 10^9$ = 1.148\ 	ext{ppb}$$

Question 13

A 11.76 mL sample of drinking water contains 0.000400 μg of lead. What is the concentration of lead in ppt? (Assume density of water is 1.00 g/mL)

Accepted Answer

$$\frac{0.000400 	imes 10$$^{-6}$$}{11.76 	imes 10$$^{-3}$$} 	imes 10$$^{12}$$ = 34.0\ 	ext{ppt}$$

Question 14

The concentration of the alkane C$$_{21}H$_{44} ($$molar mass = 296.5 g/mol) in natural water is $$8.90 	imes 10$$^{-9}$$ M. Convert this concentration to ppb. (Assume density of water is 1.00 g/mL)

Accepted Answer

$$8.90 	imes 10$$^{-9}$$ 	imes 296.5 	imes 10^3$ = 2.64\ 	ext{ppb}$$

Question 15

Which equation is used to calculate the volume of concentrated solution needed to prepare a dilute solution of desired molarity?

Accepted Answer

$$M_{	ext{conc}} V_{	ext{conc}} = M_{	ext{dilute}} V_{	ext{dilute}}$$

Question 16

You are asked to prepare 2.00 L of 0.968 M HCl solution from 12.1 M concentrated HCl. What volume of concentrated HCl is required?

Accepted Answer

$$V_{	ext{conc}} = \frac{0.968 	imes 2.00}{12.1} = 0.160\ 	ext{L} = 160\ 	ext{mL}$$

Question 17

Which of the following best describes the correct procedure for preparing a dilute acid solution from a concentrated stock?

Accepted Answer

Add the measured volume of concentrated acid to water slowly, then dilute to the final volume and mix well.

Concentration Units in Chemistry: Definitions, Calculations, and Applications

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