Q: A solution is prepared by dissolving 8.0 g of NaCl in 100 g of water. What is the percent by mass of NaCl in the solution?

$\frac{8.0}{8.0 + 100} \times 100 = 7.41\%$

Q: What is the mass of glucose present in 250 g of a 7.1% (by mass) glucose solution?

$250 \times 0.071 = 17.75\ \text{g}$

Q: A student prepares a solution by dissolving 2.4 g of NaCl in 0.1 L of water. What is the molarity of the solution? (Molar mass of NaCl = 58.5 g/mol)

$\frac{2.4\ \text{g}}{58.5\ \text{g/mol}} = 0.04103\ \text{mol}$; $M = \frac{0.04103}{0.1} = 0.4103\ \text{M}$

Q: How many grams of CuSO$_4$ are needed to prepare 125 mL of a 0.625 M solution? (Molar mass of CuSO$_4$ = 159.5 g/mol)

$0.625\ \text{mol/L} \times 0.125\ \text{L} = 0.078125\ \text{mol}$; $0.078125\ \text{mol} \times 159.5\ \text{g/mol} = 12.47\ \text{g}$

Q: A solution contains 25.0 g of glucose (C$_6$H$_{12}$O$_6$) in 364.8 g of water. What is the percent by mass of glucose in the solution?

$\frac{25.0}{25.0 + 364.8} \times 100 = 6.41\%$

Q: What is the molarity of a solution containing 17.4 g of CuSO$_4$ in 0.250 L of solution? (Molar mass of CuSO$_4$ = 159.5 g/mol)

$\frac{17.4}{159.5} = 0.1091\ \text{mol}$; $M = \frac{0.1091}{0.250} = 0.436\ \text{M}$

Q: A laboratory technician needs to prepare 175 mL of a 0.25 M HCl solution from a 2.0 M stock solution. What volume of the stock solution is required?

$M_1V_1 = M_2V_2$; $V_1 = \frac{M_2V_2}{M_1} = \frac{0.25 \times 175}{2.0} = 21.875\ \text{mL}$

Q: If 0.50 L of 0.125 M NaCl solution is diluted to a final volume of 0.750 L, what is the molarity of the resulting solution?

$M_1V_1 = M_2V_2$; $M_2 = \frac{M_1V_1}{V_2} = \frac{0.125 \times 0.50}{0.750} = 0.0833\ \text{M}$

Question 1

Which of the following will NOT speed up the dissolving process of a solid solute in a liquid solvent?

Accepted Answer

Decreasing the surface area of the solute

Question 2

A solution is prepared by dissolving 8.0 g of NaCl in 100 g of water. What is the percent by mass of NaCl in the solution?

Accepted Answer

$\frac{8.0}{8.0 + 100} 	imes 100 = 7.41\%$

Question 3

What is the mass of glucose present in 250 g of a 7.1% (by mass) glucose solution?

Accepted Answer

$250 	imes 0.071 = 17.75\ 	ext{g}$

Question 4

A student prepares a solution by dissolving 2.4 g of NaCl in 0.1 L of water. What is the molarity of the solution? (Molar mass of NaCl = 58.5 g/mol)

Accepted Answer

$\frac{2.4\ 	ext{g}}{58.5\ 	ext{g/mol}} = 0.04103\ 	ext{mol}$; $M = \frac{0.04103}{0.1} = 0.4103\ 	ext{M}$

Question 5

How many grams of CuSO$_4$ are needed to prepare 125 mL of a 0.625 M solution? (Molar mass of CuSO$_4$ = 159.5 g/mol)

Accepted Answer

$0.625\ 	ext{mol/L} 	imes 0.125\ 	ext{L} = 0.078125\ 	ext{mol}$; $0.078125\ 	ext{mol} 	imes 159.5\ 	ext{g/mol} = 12.47\ 	ext{g}$

Question 6

A solution contains 25.0 g of glucose (C$_6$H$_{12}$O$_6$) in 364.8 g of water. What is the percent by mass of glucose in the solution?

Accepted Answer

$\frac{25.0}{25.0 + 364.8} 	imes 100 = 6.41\%$

Question 7

What is the molarity of a solution containing 17.4 g of CuSO$_4$ in 0.250 L of solution? (Molar mass of CuSO$_4$ = 159.5 g/mol)

Accepted Answer

$\frac{17.4}{159.5} = 0.1091\ 	ext{mol}$; $M = \frac{0.1091}{0.250} = 0.436\ 	ext{M}$

Question 8

A laboratory technician needs to prepare 175 mL of a 0.25 M HCl solution from a 2.0 M stock solution. What volume of the stock solution is required?

Accepted Answer

$M_1V_1 = M_2V_2$; $V_1 = \frac{M_2V_2}{M_1} = \frac{0.25 	imes 175}{2.0} = 21.875\ 	ext{mL}$

Question 9

If 0.50 L of 0.125 M NaCl solution is diluted to a final volume of 0.750 L, what is the molarity of the resulting solution?

Accepted Answer

$M_1V_1 = M_2V_2$; $M_2 = \frac{M_1V_1}{V_2} = \frac{0.125 	imes 0.50}{0.750} = 0.0833\ 	ext{M}$

Question 10

Which of the following best explains why stirring a solution speeds up the dissolving process?

Accepted Answer

Stirring spreads the solute particles throughout the solvent, increasing contact between solute and solvent

Question 11

A chemist needs to prepare 50.4 mL of a 0.25 M HCl solution from a 2.0 M stock solution. What volume of stock solution should be used?

Accepted Answer

$V_1 = \frac{M_2V_2}{M_1} = \frac{0.25 	imes 50.4}{2.0} = 6.3\ 	ext{mL}$

Question 12

Which of the following is the correct formula for calculating molarity?

Accepted Answer

$M = \frac{	ext{moles of solute}}{	ext{liters of solution}}$

Question 13

A solution is made by dissolving 25 g of NaCl in 175 g of water. What is the percent by mass of NaCl in the solution?

Accepted Answer

$\frac{25}{25 + 175} 	imes 100 = 12.5\%$

Question 14

Which of the following statements about surface area and dissolving rate is correct?

Accepted Answer

Increasing the surface area of the solute increases the rate of dissolving

Question 15

A solution is prepared by dissolving 10 g of KOH in enough water to make 250 mL of solution. What is the molarity of the solution? (Molar mass of KOH = 56.1 g/mol)

Accepted Answer

$\frac{10}{56.1} = 0.1782\ 	ext{mol}$; $M = \frac{0.1782}{0.250} = 0.713\ 	ext{M}$

Question 16

Which of the following is a correct application of the dilution equation $M_1V_1 = M_2V_2$?

Accepted Answer

To find the volume of stock solution needed to prepare a solution of lower concentration

Solutions and Concentration Calculations: Study Notes for Introductory Chemistry

Study Guide - Practice Questions

Study Guide - Flashcards