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To solve this problem, we need to calculate the mass of zinc nitrate (\( \text{Zn(NO}_3\text{)}_2 \)) added to the reaction flask, given the following information:

- Volume of solution: 80.0 mL
- Molarity of solution: 0.457 M (moles of zinc nitrate per liter of solution)

**Step 1: Convert mL to L**
We convert the volume from milliliters to liters:
\[
80.0 \, \text{mL} = 0.0800 \, \text{L}
\]

**Step 2: Calculate moles of zinc nitrate**
Using the molarity formula (\( M = \frac{n}{V} \)), where:
- \( M \) is the molarity,
- \( n \) is the number of moles,
- \( V \) is the volume in liters,

We calculate the number of moles of zinc nitrate:
\[
n = M \times V = 0.457 \, \text{mol/L} \times 0.0800 \, \text{L} = 0.03656 \, \text{mol}
\]

**Step 3: Calculate molar mass of zinc nitrate (\( \text{Zn(NO}_3\text{)}_2 \))**

The molar mass of \( \text{Zn(NO}_3\text{)}_2 \) is the sum of the atomic masses of all atoms in the compound:
- Zinc (Zn): 65.38 g/mol,
- Nitrogen (N): 14.01 g/mol,
- Oxygen (O): 16.00 g/mol.

So, the molar mass of \( \text{Zn(NO}_3\text{)}_2 \) is:
\[
\text{Molar mass} = 65.38 + 2(14.01 + 3 \times 16.00) = 189.39 \, \text{g/mol}
\]

**Step 4: Calculate the mass of zinc nitrate**

Now, we calculate the mass using the formula:
\[
\text{Mass} = \text{moles} \times \text{molar mass} = 0.03656 \, \text{mol} \times 189.39 \, \text{g/mol} = 6.924 \, \text{g}
\]

**Step 5: Apply significant figures**

The data provided (volume: 80.0 mL, molarity: 0.457 M) has 3 significant figures, so we round the mass to 3 significant figures:
\[
\boxed{6.92 \, \text{g}}
\]

The mass of zinc nitrate added is **6.92 grams**.

Would you like further details on any of the steps?

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Here are 5 related questions for deeper exploration:
1. How would the mass change if the molarity was 0.500 M instead?
2. What is the role of significant figures in ensuring accurate scientific measurements?
3. How is the molar mass of a compound determined for more complex molecules?
4. How would you adjust the calculation if the volume was in liters directly?
5. What is the difference between molarity and molality in terms of solution concentration?

**Tip**: Always keep track of your units in chemical calculations to avoid errors, especially when converting between mL and L or g and kg.

A chemist adds 80.0 mL of a 0.457 M zinc nitrate (Zn(NO3)2) solution to a reaction flask. Calculate the mass in grams of zinc nitrate the chemist has added to the flask. Be sure your answer has the correct number of significant digits.

Learn how to calculate the mass of zinc nitrate added to a reaction flask using molarity and volume. This step-by-step solution includes converting mL to L, calculating moles, determining molar mass, and applying significant figures. Suitable for high school chemistry students.

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