You have a freshly prepared 0.1 M solution of glucose in water. Each liter of this solution contains how many glucose molecules?

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**Educational Content: Calculating the Number of Glucose Molecules in a Solution**

**Problem Statement:**

You have a freshly prepared 0.1 M solution of glucose in water. Each liter of this solution contains how many glucose molecules?

**Options:**

- \(6.02 \times 10^{22}\)
- \(3.01 \times 10^{23}\)
- \(6.02 \times 10^{24}\)
- \(12.04 \times 10^{23}\)

**Explanation:**

To solve this problem, one must understand the concept of molarity and Avogadro's number. Molarity (M) is defined as the number of moles of solute per liter of solution. Here, a 0.1 M solution of glucose means there are 0.1 moles of glucose in every liter of solution. 

Avogadro's number, \(6.02 \times 10^{23}\), is the number of molecules in one mole of any substance. Thus, the number of glucose molecules in 0.1 moles can be calculated by multiplying:

\[ 
0.1 \, \text{mol} \times 6.02 \times 10^{23} \, \text{molecules/mol} = 6.02 \times 10^{22} \, \text{molecules}
\]

Therefore, the correct answer is \(6.02 \times 10^{22}\).
Transcribed Image Text:**Educational Content: Calculating the Number of Glucose Molecules in a Solution** **Problem Statement:** You have a freshly prepared 0.1 M solution of glucose in water. Each liter of this solution contains how many glucose molecules? **Options:** - \(6.02 \times 10^{22}\) - \(3.01 \times 10^{23}\) - \(6.02 \times 10^{24}\) - \(12.04 \times 10^{23}\) **Explanation:** To solve this problem, one must understand the concept of molarity and Avogadro's number. Molarity (M) is defined as the number of moles of solute per liter of solution. Here, a 0.1 M solution of glucose means there are 0.1 moles of glucose in every liter of solution. Avogadro's number, \(6.02 \times 10^{23}\), is the number of molecules in one mole of any substance. Thus, the number of glucose molecules in 0.1 moles can be calculated by multiplying: \[ 0.1 \, \text{mol} \times 6.02 \times 10^{23} \, \text{molecules/mol} = 6.02 \times 10^{22} \, \text{molecules} \] Therefore, the correct answer is \(6.02 \times 10^{22}\).
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