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Concentration review worksheet Name _______________________
How many grams of beryllium chloride are needed to make 125 g of a 22.4% solution?
How many grams of beryllium chloride would you need to add to 125 mL of water to make a 0.050 molal solution
If 10.6 grams of ethanol would be mixed with 225 mL of water what is the % mass of the solution?
Explain how to make one liter of a 1.25 molar ammonium hydroxide solution.
What is the molarity of a solution in which 0.45 grams of sodium nitrate are dissolved in 265 mL of solution.
What is the molality of the sulfuric acid in a solution made by adding 3.4 grams of sulfuric acid to 3,500 g of water?
What will the volume of a 0.50 M solution be if it contains 25 grams of calcium hydroxide?
How many grams of ammonia are present in 5.0 L of a 0.050 M solution?
If I make a solution by adding 83 grams of sodium hydroxide to 750 mL of water… a) b)
What is the molality of sodium hydroxide in this solution?
What is the percent by mass of sodium hydroxide in this solution?
If I make a solution by adding water to 35 mL of methanol (CH
3
OH) until the final volume of the solution is
275 mL… a) What is the molarity of methanol in this solution? (The density of methanol is 0.792 g/mL)

1) How many grams of beryllium chloride are needed to make 125 g of a 22.4% solution?
% mass = mass solute/mass of solution
.224 =
X g BeCl
2
125g soln
= 28.0 grams of BeCl
2
2) How many grams of beryllium chloride would you need to add to 125 mL of water to make a 0.050 molal solution?
Mol = m * kg = (.050 m) (.125 kg)
.00625 mol BeCl
2
80.0 g BeCl
2
1 1mol BeCl
2
= 0.50 grams
3) If 10.6 grams of ethanol would be mixed with 225 mL of water what is the % mass of the solution?
10.6 g ethanol/(10.6 g + 225 g) *100 = 4.5 %
4) Explain how to make one liter of a 1.25 molar ammonium hydroxide solution
Mol = M * L = (1.25 M) (1.0 L)
1.25 mol NH
4
OH 35.0 g NH
4
OH
= 43.8 g
1 1 mol NH
4
OH
Dissolve 43.8 grams (1.25 moles) of ammonium hydroxide in 1 L H
2
O.
5) What is the molarity of a solution in which 0.45 grams of sodium nitrate are dissolved in
265 mL of solution.
0.45 g NaNO
3
1 mol NaNO
3
= 0.00529 mol NaNO
3
1 85.0 g NaNO
3
(.00529 mol NaNO
3
) / (.265 L ) = 0.020 M
6) What is the molality of the sulfuric acid in a solution made by adding 3.4 grams of sulfuric acid to 3.500 kg of water?
3.4 g H
2
SO
4
1 mol H
2
SO
4
1 98.1 g H
2
SO
4
=(.0347 mol H
2
SO
4
) /( 3.5 kg) = .010 m

7) What will the volume of a 0.50 M solution be if it contains 25 grams of calcium hydroxide?
25 g Ca(OH)
2
1 mol Ca(OH)
2
1 74.1 g Ca(OH)
2
= (.337 mol Ca(OH)
2
)/ (0.50 M) = .
8) How many grams of ammonia are present in 5.0 L of a 0.050 M solution? mol of NH
3
= (0.05M )(5.0 L)= .25 mol of NH
3
0.25 mol NH
3
1
17.0 g NH
3
=
1 mol NH
3
9)
If I make a solution by adding 83 grams of sodium hydroxide to 750 mL of water…
83 g NaOH 1 mol NaOH
= 2.08 mol NaOH
1 40.0 g NaOH
a) What is the molality of sodium hydroxide in this solution?
(2.08 mol / 0.75 kg water) = 2.77 m b) What is the percent by mass of sodium hydroxide in this solution?
(83 g / (83g +750 g) or 833 g * 100%) = 9.96%
10) If I make a solution by adding water to 35 mL of methanol (CH
3
OH) until the final volume of the solution is 275 mL… a) What is the molarity of methanol in this solution? (The density of methanol is 0.792 g/mL)
35 ml CH
3
OH 0.792 g CH
3
OH 1 mole CH
3
OH
1 1 ml CH
3
OH 32.0 g CH
3
OH
= 0.866 mole CH
3
OH
M = moles/volume
M = 0.866 moles CH
3
OH = 3.15 M CH
3
OH
0.275 L