A student placed 11.5 gg of glucose (C6H12O6C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then carefully added additional water until the 100.-mLmL mark on the neck of the flask was reached. The flask was then shaken until the solution was uniform. A 45.0-mLmL sample of this glucose solution was diluted to 0.500 LL. How many grams of glucose are in 100. mLmL of the final solution?

We can approach this problem step by step using the concept of dilution and concentration.

### Step 1: Calculate the molarity of the original glucose solution

The student first prepares a glucose solution by dissolving 11.5 g of glucose in 100 mL of water. Let's calculate the molarity of this original solution.

- The molar mass of glucose (\(C_6H_{12}O_6\)) is calculated as:

\[
\text{Molar mass of glucose} = 6(12.01) + 12(1.008) + 6(16.00) = 180.16 \ \text{g/mol}
\]

Now, we calculate the number of moles of glucose in the 11.5 g sample:

\[
\text{Moles of glucose} = \frac{11.5 \ \text{g}}{180.16 \ \text{g/mol}} = 0.06383 \ \text{mol}
\]

Since the glucose is dissolved in 100 mL (0.100 L) of solution, the molarity (\(M\)) of the original solution is:

\[
M = \frac{0.06383 \ \text{mol}}{0.100 \ \text{L}} = 0.6383 \ \text{mol/L}
\]

### Step 2: Determine the moles of glucose in the 45.0 mL sample

Next, the student takes a 45.0 mL sample from this original solution. To find the moles of glucose in this sample, we use the molarity of the original solution:

\[
\text{Moles of glucose in 45.0 mL} = 0.6383 \ \text{mol/L} \times 0.0450 \ \text{L} = 0.02872 \ \text{mol}
\]

### Step 3: Determine the molarity of the diluted solution

The 45.0 mL sample is then diluted to 0.500 L. To find the new molarity of the solution after dilution, we use the formula:

\[
M_1 V_1 = M_2 V_2
\]

Where:
- \(M_1 = 0.6383 \ \text{mol/L}\) (molarity of the original solution)
- \(V_1 = 0.0450 \ \text{L}\) (volume of the original solution)
- \(M_2\) is the molarity of the diluted solution (what we need to find)
- \(V_2 = 0.500 \ \text{L}\) (final volume after dilution)

Rearranging the equation to solve for \(M_2\):

\[
M_2 = \frac{M_1 V_1}{V_2} = \frac{0.6383 \times 0.0450}{0.500} = 0.05745 \ \text{mol/L}
\]

### Step 4: Calculate the moles and mass of glucose in 100 mL of the final solution

Now, we are asked to find how many grams of glucose are in 100 mL (0.100 L) of the final diluted solution. First, we calculate the moles of glucose in this volume:

\[
\text{Moles of glucose in 100 mL} = 0.05745 \ \text{mol/L} \times 0.100 \ \text{L} = 0.005745 \ \text{mol}
\]

Finally, we convert moles to grams using the molar mass of glucose:

\[
\text{Mass of glucose} = 0.005745 \ \text{mol} \times 180.16 \ \text{g/mol} = 1.034 \ \text{g}
\]

### Final Answer:
There are **1.034 grams** of glucose in 100 mL of the final solution.

---

Do you have any questions or would you like more details on any of the steps? Here are some related questions to deepen your understanding:

1. What would happen to the glucose concentration if the solution were diluted further?
2. How would you calculate the concentration of glucose in mol/L for different solution volumes?
3. Can you explain how the molar mass of glucose is calculated from its molecular formula?
4. How does the process of dilution affect the mass of solute in a solution?
5. If the final solution were 1 L, how would the concentration and mass of glucose change?

**Tip**: Always ensure that your volumes are in liters when working with molarity (mol/L).

A student placed 11.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then carefully added additional water until the 100-mL mark on the neck of the flask was reached. The flask was then shaken until the solution was uniform. A 45.0-mL sample of this glucose solution was diluted to 0.500 L. How many grams of glucose are in 100 mL of the final solution?

Learn how to calculate the grams of glucose in a diluted solution. This problem involves molarity, dilution, and the use of the dilution equation M1V1 = M2V2. Ideal for high school students studying chemistry and focusing on molar mass and solution concentration.

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