Sure, let's solve this step-by-step.
### Given:
1. Mass of CuNO[tex]\(_3\)[/tex] = 8.62 grams
2. Molarity (M) of the solution = 0.750 M
### Step 1: Calculate the molecular weight of CuNO[tex]\(_3\)[/tex]
- Atomic weight of Copper (Cu) = 63.55 g/mol
- Atomic weight of Nitrogen (N) = 14.01 g/mol
- Atomic weight of Oxygen (O) = 16.00 g/mol
Molecular weight of CuNO[tex]\(_3\)[/tex] = 63.55 + 14.01 + (16.00 * 3)
= 63.55 + 14.01 + 48.00
= 125.56 g/mol
### Step 2: Calculate the moles of CuNO[tex]\(_3\)[/tex]
To find the moles of CuNO[tex]\(_3\)[/tex], use the formula:
[tex]\[
\text{{moles}} = \frac{{\text{{mass}}}}{{\text{{molecular weight}}}}
\][/tex]
[tex]\[
\text{{moles of CuNO}}_3 = \frac{{8.62 \, \text{g}}}{{125.56 \, \text{g/mol}}} \approx 0.06865 \, \text{moles}
\][/tex]
### Step 3: Calculate the volume of the solution in liters
To find the volume of the solution in liters, use the relationship between moles, molarity, and volume:
[tex]\[
\text{{molarity}} = \frac{{\text{{moles}}}}{{\text{{volume in liters}}}}
\][/tex]
Rearranging for volume in liters:
[tex]\[
\text{{volume in liters}} = \frac{{\text{{moles}}}}{{\text{{molarity}}}}
\][/tex]
[tex]\[
\text{{volume in liters}} = \frac{{0.06865 \, \text{moles}}}{{0.750 \, \text{M}}} \approx 0.09154 \, \text{liters}
\][/tex]
### Step 4: Convert the volume to milliliters
1 liter = 1000 milliliters
[tex]\[
\text{{volume in milliliters}} = 0.09154 \, \text{liters} \times 1000 \, \text{mL/L} \approx 91.54 \, \text{mL}
\][/tex]
### Final Answer:
The volume of the solution is approximately [tex]\(91.54\)[/tex] milliliters.
