Question

1. how many liters of a 0.88 m solution can be made with 25.5 grams of lithium fluoride?
2. what is the concentration of a solution with a volume of 660 ml that contains 33.4 grams of aluminum acetate?
3. how many liters of a 0.75 m solution can be made with 75 grams of lead (ii) oxide?
4. how many grams of manganese (iv) oxide are needed to make 5.6 liters of a 2.1 m solution?
5. what is the concentration of a solution with a volume of 9 ml that contains 2 grams of iron (iii) hydroxide?
6. how many liters of a 3.4 m isopropanol solution can be made with 78 grams of isopropanol ($c_3h_8o$)?
7. what is the concentration of a solution with a volume 3.3 ml that contains 12 grams of ammonium sulfite?

Explanation:

Problem 7

Step1: Calculate molar mass of LiF

Molar mass of LiF: $M(\text{LiF}) = 6.94 + 19.00 = 25.94\ \text{g/mol}$

Step2: Find moles of LiF

$n(\text{LiF}) = \frac{m}{M} = \frac{25.5}{25.94} \approx 0.983\ \text{mol}$

Step3: Calculate volume of solution

$V = \frac{n}{M} = \frac{0.983}{0.88} \approx 1.12\ \text{L}$

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Problem 8

Step1: Molar mass of $\text{Al(C}_2\text{H}_3\text{O}_2\text{)}_3$

$M(\text{Al(C}_2\text{H}_3\text{O}_2\text{)}_3) = 26.98 + 3\times(2\times12.01+3\times1.01+2\times16.00) = 204.11\ \text{g/mol}$

Step2: Moles of aluminum acetate

$n = \frac{33.4}{204.11} \approx 0.164\ \text{mol}$

Step3: Convert volume to liters

$V = 660\ \text{mL} = 0.660\ \text{L}$

Step4: Calculate concentration

$M = \frac{n}{V} = \frac{0.164}{0.660} \approx 0.248\ \text{M}$

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Problem 9

Step1: Molar mass of PbO

$M(\text{PbO}) = 207.2 + 16.00 = 223.2\ \text{g/mol}$

Step2: Moles of PbO

$n = \frac{75}{223.2} \approx 0.336\ \text{mol}$

Step3: Calculate solution volume

$V = \frac{n}{M} = \frac{0.336}{0.75} \approx 0.448\ \text{L}$

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Problem 10

Step1: Molar mass of $\text{MnO}_2$

$M(\text{MnO}_2) = 54.94 + 2\times16.00 = 86.94\ \text{g/mol}$

Step2: Moles of $\text{MnO}_2$ needed

$n = M\times V = 2.1\times5.6 = 11.76\ \text{mol}$

Step3: Calculate mass of $\text{MnO}_2$

$m = n\times M = 11.76\times86.94 \approx 1022\ \text{g}$

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Problem 11

Step1: Molar mass of $\text{Fe(OH)}_3$

$M(\text{Fe(OH)}_3) = 55.85 + 3\times(16.00+1.01) = 106.88\ \text{g/mol}$

Step2: Moles of $\text{Fe(OH)}_3$

$n = \frac{2}{106.88} \approx 0.0187\ \text{mol}$

Step3: Convert volume to liters

$V = 9\ \text{mL} = 0.009\ \text{L}$

Step4: Calculate concentration

$M = \frac{n}{V} = \frac{0.0187}{0.009} \approx 2.08\ \text{M}$

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Problem 12

Step1: Molar mass of $\text{C}_3\text{H}_8\text{O}$

$M(\text{C}_3\text{H}_8\text{O}) = 3\times12.01 + 8\times1.01 + 16.00 = 60.11\ \text{g/mol}$

Step2: Moles of isopropanol

$n = \frac{78}{60.11} \approx 1.30\ \text{mol}$

Step3: Calculate solution volume

$V = \frac{n}{M} = \frac{1.30}{3.4} \approx 0.382\ \text{L}$

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Problem 13

Step1: Molar mass of $(\text{NH}_4)_2\text{SO}_3$

$M((\text{NH}_4)_2\text{SO}_3) = 2\times(14.01+4\times1.01) + 32.07 + 3\times16.00 = 116.15\ \text{g/mol}$

Step2: Moles of ammonium sulfite

$n = \frac{12}{116.15} \approx 0.103\ \text{mol}$

Step3: Convert volume to liters

$V = 3.3\ \text{mL} = 0.0033\ \text{L}$

Step4: Calculate concentration

$M = \frac{n}{V} = \frac{0.103}{0.0033} \approx 31.2\ \text{M}$

Answer:

1. $\approx 1.12$ liters
2. $\approx 0.248$ M
3. $\approx 0.448$ liters
4. $\approx 1022$ grams
5. $\approx 2.08$ M
6. $\approx 0.382$ liters
7. $\approx 31.2$ M