50 mL of an aqueous solution of glucose `C_(6)H_(12)o_(6)` (Molar mass : 180 g/mol) contains `6.02 xx10^(22)` molecules. The concentration of the solution will be?

Options

A. 0.1 M
B. 0.2 M
C. 1.0 M
D. 2.0M

Explanation

To find the concentration of the glucose solution, we need to calculate the molarity (M) of the solution. Molarity is defined as the number of moles of solute per liter of solution.

### Step 1: Calculate the number of moles of glucose
We know that:
- The number of molecules of glucose in the solution = \(6.02 \times 10^{22}\)
- Avogadro's number = \(6.022 \times 10^{23}\) molecules/mole

To find the number of moles (n) of glucose, we can use the formula:
\[ n = \frac{\text{Number of molecules}}{\text{Avogadro's number}} \]
Substituting the values:
\[ n = \frac{6.02 \times 10^{22}}{6.022 \times 10^{23}} \approx 0.1 \text{ moles} \]

### Step 2: Convert the volume of the solution from mL to L
The volume of the solution is given as 50 mL. To convert this to liters:
\[ \text{Volume in liters} = \frac{50 \text{ mL}}{1000} = 0.05 \text{ L} \]

### Step 3: Calculate the molarity of the solution
Molarity (M) is calculated using the formula:
\[ M = \frac{\text{Number of moles of solute}}{\text{Volume of solution in liters}} \]
Substituting the values we found:
\[ M = \frac{0.1 \text{ moles}}{0.05 \text{ L}} = 2 \text{ M} \]

### Final Answer
The concentration of the glucose solution is **2 M**.

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