To determine the volume of a 1.0 M (molar) C₆H₁₂O₆ (glucose) solution that can be made using 360.4 grams of C₆H₁₂O₆, follow these steps:
1. First, calculate the number of moles of glucose using the given mass and the molar mass.
[tex]\[
\text{Number of moles} = \frac{\text{Mass of solute}}{\text{Molar mass of solute}}
\][/tex]
Given:
- Mass of solute (C₆H₁₂O₆) = 360.4 grams
- Molar mass of solute (C₆H₁₂O₆) = 180.2 grams per mole
Substituting the values:
[tex]\[
\text{Number of moles} = \frac{360.4\ \text{grams}}{180.2\ \text{grams per mole}} = 2.0\ \text{moles}
\][/tex]
2. Next, use the molarity formula to find the volume in liters. Molarity (M) is defined as the number of moles of solute per liter of solution.
[tex]\[
\text{Molarity} = \frac{\text{Number of moles}}{\text{Volume in liters}}
\][/tex]
Rearranging to solve for volume:
[tex]\[
\text{Volume in liters} = \frac{\text{Number of moles}}{\text{Molarity}}
\][/tex]
Given:
- Molarity (M) = 1.0 M
- Number of moles = 2.0 moles
Substituting the values:
[tex]\[
\text{Volume in liters} = \frac{2.0\ \text{moles}}{1.0\ \text{M}} = 2.0\ \text{liters}
\][/tex]
Therefore, you can make 2.0 liters of a 1.0 M C₆H₁₂O₆ solution using 360.4 grams of glucose.
