Name: ______________________________________________ Date: __________________ Pd: _____
Practice Problems - Solution Chemistry – to address Section 1.5 in Green and Damji – pp 26-29
1.
Find the concentration of a solution containing 2.00 g of sodium hydroxide in 125 cm 3 of solution.
2.
What amount of sulfuric acid is present in 25.0 cm 3 of 2.00 M sulfuric acid, H
2
SO
4
?
3.
Calculate the amount of hydrochloric acid present in 150 cm 3 of 1.35 mol dm -3 HCl (aq).
4.
What mass of potassium permanganate is present in 25.55 cm 3 of 0.35 mol dm -3 KMnO4 (aq)?
5.
What volume (in dm 3 ) of 0.30 mol dm -3 NaNO
3
solution can be prepared from 0.060 mol of solute?
6.
How would you prepare 250 cm 3 of a 0.200 mol dm -3 solution of sodium chloride?
7.
How would you make 250. mL of a 3M HNO
3
solution from concentrated (15.8M) HNO
3
?
8.
What will be the concentration of the solution formed by mixing 200 cm 3 of 3.00 mol dm -3 HCl (aq) with 300 cm 3 of 1.50 mol dm -3 HCl?
9.
Which of the following solutions contains the greatest amount of chloride ions? a.
10.0 cm 3 of 0.250 M NaCl
(aq) b.
10.0 cm 3 of 0.250 M CaCl
2 (aq) c.
20.0 cm 3 of 0.250 M NaCl
(aq) d.
20.0 cm 3 of 0.250 M CaCl
2 (aq) e.
10.0 cm 3 of 0.500 M NaCl
(aq)
10.
Sodium reacts with water as follows:
2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g)
1.15 g of sodium is allowed to react completely with water. The resulting solution is diluted to
250 cm3. Calculate the concentration, in mol dm–3, of the resulting sodium hydroxide solution.
11.
What volume of 0.500 mol dm−3 HCl(aq) is required to react completely with 10.0 g of calcium carbonate according to the equation below?
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)

KEY to Practice Problems - Solution Chemistry – to address Section 1.5 in Green and Damji – pp 26-29
1.
Find the concentration of a solution containing 2.00 g of sodium hydroxide in 125 cm 3 of solution.
Strategy: Remember C = n/V [C=concentration in mol dm -3 or mol/L; n=mol; V=volume in liters]
(a) determine the moles of sodium hydroxide – you’ll need to figure out the formula for sodium hydroxide and calculate the molar mass of sodium hydroxide
(b) convert volume into liters
(c) calculate concentration by dividing the number of moles by the volume
Answer: 0.400 M
2.
What amount of sulfuric acid is present in 25.0 cm 3 of 2.00 M sulfuric acid, H
2
SO
4
?
Strategy: Just rearrange the equation C = n/V to n = C*V
(a) convert volume to liters (b) divide the concentration by the volume
Answer: 0.0500 mol
3.
Calculate the amount of hydrochloric acid present in 150 cm 3 of 1.35 mol dm -3 HCl (aq).
Strategy: see # 2
Answer: 0.203 mol
4.
What mass of potassium permanganate is present in 25.55 cm 3 of 0.35 mol dm -3 KMnO4 (aq)?
Strategy: Solve for amount, using n = C*V (just like #2 and #3). Once you have the amount (moles) you can convert to mass using the molar mass mass = moles/molar mass
Answer: 1.41 g
5.
What volume (in dm 3 ) of 0.30 mol dm -3 NaNO
3
solution can be prepared from 0.060 mol of solute?
Strategy: Solve for volume, using V = n/C which is just another rearrangement of C = n/V
Answer: 0.20 dm 3
6.
How would you prepare 250 cm 3 of a 0.200 mol dm -3 solution of sodium chloride?
Strategy: Determine the amount of sodium chloride you would need, using the relationship n = C*V .
Be sure your volume is in liters. Then convert the number of moles to grams – since sodium chloride is a solid.
Then describe how you would actually make the solution.
Answer: 2.922 g NaCl. Measure out 2.922 grams of sodium chloride in a small beaker. Add about 100 mL of distilled water to the sodium chloride and mix completely. Use a funnel to pour this solution into a 250 mL volumetric flask. Use additional distilled water to rinse the beaker, pouring this rinse water into the flask. Mix completely and repeat several times.
Finally, add distilled water to the etched line on the flask. Mix completely.
7.
How would you make 250. mL of a 3M HNO
3
solution from concentrated (15.8M) HNO
3
?
Strategy: Use the relationship C
1
V
1
= C
2
V
2
to solve for the volume of concentrated acid you need. Then describe how you would actually make the solution.
Answer: Get a 250 mL volumetric flask and add about 150-175 mL of distilled water to the flask. Carefully measure out 47.5 mL of concentrated nitric acid. Carefully pour the concentrated acid into the volumetric flask, mixing as you proceed. Once the concentrated acid has been transferred into the flask, add more distilled water until the level reaches the etched mark. Stopper the flask and mix completely.

8.
What will be the concentration of the solution formed by mixing 200 cm 3 of 3.00 mol dm -3 HCL (aq) with 300 cm 3 of 1.50 mol dm -3 HCl?
Strategy: Determine the number of moles of HCl in each solution, then add these two values together to get the total number of moles of HCl. Determine the total volume of the resulting solution (in L) by adding the two volumes together. Use the equation C = n/V to solve for the resulting concentration.
Answer: 2.10 mol dm -3
9.
Which of the following solutions contains the greatest amount of chloride ions? a.
10.0 cm 3 of 0.250 M NaCl
(aq) b.
10.0 cm 3 of 0.250 M CaCl
2 (aq) c.
20.0 cm 3 of 0.250 M NaCl
(aq) d.
20.0 cm 3 of 0.250 M CaCl
2 (aq) e.
10.0 cm 3 of 0.500 M NaCl
(aq)
Strategy: Determine the amount of each salt, using the relationship n = C*V .
Be sure your volume is in liters. Then convert the number of moles of salt to number of moles of ions by looking at the molar ratio – in this case, the subscript on the chloride ion.
Answer: D - 20.0 cm 3 of 0.250 M CaCl
2 (aq)
10.
Sodium reacts with water as follows:
2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g)
1.15 g of sodium is allowed to react completely with water. The resulting solution is diluted to
250 cm3. Calculate the concentration, in mol dm–3, of the resulting sodium hydroxide solution.
Strategy: This is a stoichiometry problem. Convert the g of sodium to moles of sodium. Then multiply by the appropriate mole ratio to convert to moles of sodium hydroxide. Then use the relationship C = n/V to solve for the concentration. Remember to put your volume in dm 3 .
Answer: D – 0.20 mol dm -3
11.
What volume of 0.500 mol dm−3 HCl(aq) is required to react completely with 10.0 g of calcium carbonate according to the equation below?
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
Strategy: This is a stoichiometry problem. Convert the g of calcium carbonate to moles. Then multiply by the appropriate mole ratio to convert to moles of HCl. Then use the relationship V = n/C to solve for the volume..
Answer: 0.400 dm 3 (400. cm 3 or 400. mL)