Molarity Practice II worksheet with ten chemistry problems involving calculations of molarity, grams, and atoms in different chemical solutions.

Problem Set: Molarity Practice II

Below, I will solve each problem step-by-step with clear explanations.

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#### 1. How many grams are in 250 mL of a 0.150 M solution of AgNO₃?

Solution:
To find the mass of solute (AgNO₃) in the solution, we use the formula:
\[
\text{Mass} = \text{Moles} \times \text{Molar Mass}
\]
First, calculate the moles of AgNO₃:
\[
\text{Moles} = \text{Molarity} \times \text{Volume (in liters)}
\]
Given:
- Molarity = 0.150 M
- Volume = 250 mL = 0.250 L

\[
\text{Moles of AgNO₃} = 0.150 \, \text{M} \times 0.250 \, \text{L} = 0.0375 \, \text{moles}
\]

Next, calculate the molar mass of AgNO₃:
- Ag: 107.87 g/mol
- N: 14.01 g/mol
- O: 16.00 g/mol × 3 = 48.00 g/mol

\[
\text{Molar Mass of AgNO₃} = 107.87 + 14.01 + 48.00 = 169.88 \, \text{g/mol}
\]

Now, calculate the mass:
\[
\text{Mass} = \text{Moles} \times \text{Molar Mass} = 0.0375 \, \text{moles} \times 169.88 \, \text{g/mol} = 6.37 \, \text{g}
\]

Answer:
\[
\boxed{6.37 \, \text{g}}
\]

---

#### 2. Determine the molarity of NaOH if 28.00 grams of NaOH is dissolved to a total volume of 250 mL.

Solution:
To find the molarity, we use the formula:
\[
\text{Molarity} = \frac{\text{Moles of Solute}}{\text{Volume (in liters)}}
\]
First, calculate the moles of NaOH:
\[
\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}
\]
Given:
- Mass = 28.00 g
- Molar Mass of NaOH = 22.99 (Na) + 16.00 (O) + 1.01 (H) = 40.00 g/mol

\[
\text{Moles of NaOH} = \frac{28.00 \, \text{g}}{40.00 \, \text{g/mol}} = 0.700 \, \text{moles}
\]

Next, calculate the molarity:
\[
\text{Volume} = 250 \, \text{mL} = 0.250 \, \text{L}
\]

\[
\text{Molarity} = \frac{0.700 \, \text{moles}}{0.250 \, \text{L}} = 2.80 \, \text{M}
\]

Answer:
\[
\boxed{2.80 \, \text{M}}
\]

---

#### 3. How many formula units are present in 500 mL of a 0.250 M solution of NaCl?

Solution:
To find the number of formula units, we first calculate the moles of NaCl and then convert to formula units using Avogadro's number (\(6.022 \times 10^{23}\) formula units/mol).

First, calculate the moles of NaCl:
\[
\text{Moles} = \text{Molarity} \times \text{Volume (in liters)}
\]
Given:
- Molarity = 0.250 M
- Volume = 500 mL = 0.500 L

\[
\text{Moles of NaCl} = 0.250 \, \text{M} \times 0.500 \, \text{L} = 0.125 \, \text{moles}
\]

Next, convert moles to formula units:
\[
\text{Formula Units} = \text{Moles} \times \text{Avogadro's Number}
\]

\[
\text{Formula Units} = 0.125 \, \text{moles} \times 6.022 \times 10^{23} \, \text{formula units/mol} = 7.53 \times 10^{22} \, \text{formula units}
\]

Answer:
\[
\boxed{7.53 \times 10^{22}}
\]

---

#### 4. How many grams are in 1.50 L of a 0.500 M solution of copper(II) sulfate pentahydrate, CuSO₄·5H₂O?

Solution:
To find the mass of CuSO₄·5H₂O, we use the formula:
\[
\text{Mass} = \text{Moles} \times \text{Molar Mass}
\]
First, calculate the moles of CuSO₄·5H₂O:
\[
\text{Moles} = \text{Molarity} \times \text{Volume (in liters)}
\]
Given:
- Molarity = 0.500 M
- Volume = 1.50 L

\[
\text{Moles of CuSO₄·5H₂O} = 0.500 \, \text{M} \times 1.50 \, \text{L} = 0.750 \, \text{moles}
\]

Next, calculate the molar mass of CuSO₄·5H₂O:
- Cu: 63.55 g/mol
- S: 32.07 g/mol
- O: 16.00 g/mol × 9 = 144.00 g/mol
- H: 1.01 g/mol × 10 = 10.10 g/mol

\[
\text{Molar Mass of CuSO₄·5H₂O} = 63.55 + 32.07 + 144.00 + 10.10 = 249.72 \, \text{g/mol}
\]

Now, calculate the mass:
\[
\text{Mass} = \text{Moles} \times \text{Molar Mass} = 0.750 \, \text{moles} \times 249.72 \, \text{g/mol} = 187.29 \, \text{g}
\]

Answer:
\[
\boxed{187.3 \, \text{g}}
\]

---

#### 5. Determine the molarity of Na₂CO₃ if 16.00 grams of sodium carbonate is dissolved to a total volume of 750 mL.

Solution:
To find the molarity, we use the formula:
\[
\text{Molarity} = \frac{\text{Moles of Solute}}{\text{Volume (in liters)}}
\]
First, calculate the moles of Na₂CO₃:
\[
\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}
\]
Given:
- Mass = 16.00 g
- Molar Mass of Na₂CO₃ = 2(22.99) + 12.01 + 3(16.00) = 46.98 + 12.01 + 48.00 = 106.99 g/mol

\[
\text{Moles of Na₂CO₃} = \frac{16.00 \, \text{g}}{106.99 \, \text{g/mol}} = 0.1496 \, \text{moles}
\]

Next, calculate the molarity:
\[
\text{Volume} = 750 \, \text{mL} = 0.750 \, \text{L}
\]

\[
\text{Molarity} = \frac{0.1496 \, \text{moles}}{0.750 \, \text{L}} = 0.1995 \, \text{M}
\]

Answer:
\[
\boxed{0.200 \, \text{M}}
\]

---

#### 6. How many atoms of oxygen are present in 31.50 grams of CO₂?

Solution:
To find the number of oxygen atoms, we first calculate the moles of CO₂ and then determine the number of oxygen atoms.

First, calculate the moles of CO₂:
\[
\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}
\]
Given:
- Mass = 31.50 g
- Molar Mass of CO₂ = 12.01 (C) + 2(16.00) = 12.01 + 32.00 = 44.01 g/mol

\[
\text{Moles of CO₂} = \frac{31.50 \, \text{g}}{44.01 \, \text{g/mol}} = 0.7157 \, \text{moles}
\]

Each molecule of CO₂ contains 2 oxygen atoms. Therefore, the total number of oxygen atoms is:
\[
\text{Number of Oxygen Atoms} = \text{Moles of CO₂} \times 2 \times \text{Avogadro's Number}
\]

\[
\text{Number of Oxygen Atoms} = 0.7157 \, \text{moles} \times 2 \times 6.022 \times 10^{23} \, \text{atoms/mol} = 8.61 \times 10^{23} \, \text{atoms}
\]

Answer:
\[
\boxed{8.61 \times 10^{23}}
\]

---

#### 7. How many CO₂ molecules are present in a 2.00-liter coke? The molarity of CO₂ in coke is 1.20 × 10⁻⁴ M.

Solution:
To find the number of CO₂ molecules, we first calculate the moles of CO₂ and then convert to molecules using Avogadro's number.

First, calculate the moles of CO₂:
\[
\text{Moles} = \text{Molarity} \times \text{Volume (in liters)}
\]
Given:
- Molarity = 1.20 × 10⁻⁴ M
- Volume = 2.00 L

\[
\text{Moles of CO₂} = 1.20 \times 10^{-4} \, \text{M} \times 2.00 \, \text{L} = 2.40 \times 10^{-4} \, \text{moles}
\]

Next, convert moles to molecules:
\[
\text{Molecules} = \text{Moles} \times \text{Avogadro's Number}
\]

\[
\text{Molecules} = 2.40 \times 10^{-4} \, \text{moles} \times 6.022 \times 10^{23} \, \text{molecules/mol} = 1.45 \times 10^{20} \, \text{molecules}
\]

Answer:
\[
\boxed{1.45 \times 10^{20}}
\]

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#### 8. Describe the steps needed to make 250 mL of 0.100 M of AgNO₃.

Solution:
To prepare a 0.100 M solution of AgNO₃ in 250 mL, follow these steps:

1. Calculate the moles of AgNO₃ needed:
\[
\text{Moles} = \text{Molarity} \times \text{Volume (in liters)}
\]
Given:
- Molarity = 0.100 M
- Volume = 250 mL = 0.250 L

\[
\text{Moles of AgNO₃} = 0.100 \, \text{M} \times 0.250 \, \text{L} = 0.0250 \, \text{moles}
\]

2. Calculate the mass of AgNO₃ required:
\[
\text{Mass} = \text{Moles} \times \text{Molar Mass}
\]
Molar Mass of AgNO₃ = 107.87 + 14.01 + 48.00 = 169.88 g/mol

\[
\text{Mass of AgNO₃} = 0.0250 \, \text{moles} \times 169.88 \, \text{g/mol} = 4.25 \, \text{g}
\]

3. Weigh out 4.25 grams of AgNO₃.

4. Dissolve the 4.25 grams of AgNO₃ in a small amount of distilled water in a volumetric flask.

5. Add more distilled water to the volumetric flask until the total volume reaches 250 mL.

6. Mix thoroughly to ensure complete dissolution and uniform concentration.

Answer:
\[
\boxed{\text{Weigh 4.25 g of AgNO₃, dissolve in water, and dilute to 250 mL in a volumetric flask.}}
\]

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#### 9. The Joker (from Batman) was injured using an 18.1 molar solution of sulfuric acid (H₂SO₄). How many grams must be dissolved if the Joker fell into a 2000-Liter container?

Solution:
To find the mass of H₂SO₄ required, we use the formula:
\[
\text{Mass} = \text{Moles} \times \text{Molar Mass}
\]
First, calculate the moles of H₂SO₄:
\[
\text{Moles} = \text{Molarity} \times \text{Volume (in liters)}
\]
Given:
- Molarity = 18.1 M
- Volume = 2000 L

\[
\text{Moles of H₂SO₄} = 18.1 \, \text{M} \times 2000 \, \text{L} = 36200 \, \text{moles}
\]

Next, calculate the molar mass of H₂SO₄:
- H: 1.01 g/mol × 2 = 2.02 g/mol
- S: 32.07 g/mol
- O: 16.00 g/mol × 4 = 64.00 g/mol

\[
\text{Molar Mass of H₂SO₄} = 2.02 + 32.07 + 64.00 = 98.09 \, \text{g/mol}
\]

Now, calculate the mass:
\[
\text{Mass} = \text{Moles} \times \text{Molar Mass} = 36200 \, \text{moles} \times 98.09 \, \text{g/mol} = 3.55 \times 10^6 \, \text{g}
\]

Convert grams to kilograms:
\[
3.55 \times 10^6 \, \text{g} = 3550 \, \text{kg}
\]

Answer:
\[
\boxed{3550 \, \text{kg}}
\]

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#### 10. A tablespoon of baking soda (sodium bicarbonate) is used for a Martha Stewart cake recipe. The sodium bicarbonate massed out at 10.62 grams. What would the molarity of the solution be if Martha wanted to dissolve the 10.62 grams into 525 mL of volume (you know, for fun)?

Solution:
To find the molarity, we use the formula:
\[
\text{Molarity} = \frac{\text{Moles of Solute}}{\text{Volume (in liters)}}
\]
First, calculate the moles of NaHCO₃:
\[
\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}
\]
Given:
- Mass = 10.62 g
- Molar Mass of NaHCO₃ = 22.99 (Na) + 1.01 (H) + 12.01 (C) + 3(16.00) (O) = 22.99 + 1.01 + 12.01 + 48.00 = 84.01 g/mol

\[
\text{Moles of NaHCO₃} = \frac{10.62 \, \text{g}}{84.01 \, \text{g/mol}} = 0.1264 \, \text{moles}
\]

Next, calculate the molarity:
\[
\text{Volume} = 525 \, \text{mL} = 0.525 \, \text{L}
\]

\[
\text{Molarity} = \frac{0.1264 \, \text{moles}}{0.525 \, \text{L}} = 0.241 \, \text{M}
\]

Answer:
\[
\boxed{0.241 \, \text{M}}
\]

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Final Answers:

1. \(\boxed{6.37 \, \text{g}}\)
2. \(\boxed{2.80 \, \text{M}}\)
3. \(\boxed{7.53 \times 10^{22}}\)
4. \(\boxed{187.3 \, \text{g}}\)
5. \(\boxed{0.200 \, \text{M}}\)
6. \(\boxed{8.61 \times 10^{23}}\)
7. \(\boxed{1.45 \times 10^{20}}\)
8. \(\boxed{\text{Weigh 4.25 g of AgNO₃, dissolve in water, and dilute to 250 mL in a volumetric flask.}}\)
9. \(\boxed{3550 \, \text{kg}}\)
10. \(\boxed{0.241 \, \text{M}}\)