Alright, let's solve this step by step:
### Step 1: Calculate the Molar Mass of C[tex]\(_6\)[/tex]H[tex]\(_{12}\)[/tex]O[tex]\(_6\)[/tex]
First, we need to determine the molar mass (molecular weight) of glucose (C[tex]\(_6\)[/tex]H[tex]\(_{12}\)[/tex]O[tex]\(_6\)[/tex]).
- Carbon (C): [tex]\(6 \text{ atoms} \times 12.01 \text{ g/mol} = 72.06 \text{ g/mol}\)[/tex]
- Hydrogen (H): [tex]\(12 \text{ atoms} \times 1.01 \text{ g/mol} = 12.12 \text{ g/mol}\)[/tex]
- Oxygen (O): [tex]\(6 \text{ atoms} \times 16.00 \text{ g/mol} = 96.00 \text{ g/mol}\)[/tex]
So, the molar mass of C[tex]\(_6\)[/tex]H[tex]\(_{12}\)[/tex]O[tex]\(_6\)[/tex] is:
[tex]\[ 72.06 + 12.12 + 96.00 = 180.18 \text{ g/mol} \][/tex]
### Step 2: Calculate the Moles of C[tex]\(_6\)[/tex]H[tex]\(_{12}\)[/tex]O[tex]\(_6\)[/tex]
Next, we need to find the number of moles of glucose using the formula:
[tex]\[ \text{Moles} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}} \][/tex]
Given the mass of glucose is 32.3 grams:
[tex]\[ \text{Moles of C}_6\text{H}_{12}\text{O}_6 = \frac{32.3 \text{ g}}{180.18 \text{ g/mol}} \approx 0.179 \text{ moles} \][/tex]
### Step 3: Convert Volume of Solution from mL to L
Since molarity is defined as moles per liter, we need to convert 660 mL to liters:
[tex]\[ 660 \text{ mL} = \frac{660}{1000} = 0.66 \text{ L} \][/tex]
### Step 4: Calculate Molarity
Finally, molarity (M) of the solution is given by:
[tex]\[ \text{Molarity} = \frac{\text{Moles of Solute}}{\text{Volume of Solution in Liters}} \][/tex]
So,
[tex]\[ \text{Molarity} = \frac{0.179 \text{ moles}}{0.66 \text{ L}} \approx 0.272 \text{ M} \][/tex]
### Summary
- The molar mass of C[tex]\(_6\)[/tex]H[tex]\(_{12}\)[/tex]O[tex]\(_6\)[/tex] is 180.18 g/mol.
- The moles of C[tex]\(_6\)[/tex]H[tex]\(_{12}\)[/tex]O[tex]\(_6\)[/tex] in 32.3 grams of the substance are approximately 0.179.
- The volume of the solution is 0.66 L.
- The molarity of the solution is approximately 0.272 M.
Thus, the molarity of the 32.3 g glucose (C[tex]\(_6\)[/tex]H[tex]\(_{12}\)[/tex]O[tex]\(_6\)[/tex]) in 660 mL solution is [tex]\( \approx 0.272 \text{ M} \)[/tex].
