Two glucose solution are mixed. One has a volume of `480 mL` and a c oncentration of `1.50 M` and the second has a volume of `250 mL` and concentration `1.20 M`. The molarity of final solution is

Options

A. `1.20 M`
B. `1.50 M`
C. `1.344 M`
D. `2.70 M`

Explanation

To find the molarity of the final solution after mixing two glucose solutions, we can follow these steps:

### Step 1: Calculate the number of moles of glucose in each solution.

For the first solution:
- Volume (V1) = 480 mL = 0.480 L (convert mL to L by dividing by 1000)
- Concentration (M1) = 1.50 M

Number of moles (n1) = M1 × V1
\[ n_1 = 1.50 \, \text{mol/L} \times 0.480 \, \text{L} = 0.720 \, \text{mol} \]

For the second solution:
- Volume (V2) = 250 mL = 0.250 L
- Concentration (M2) = 1.20 M

Number of moles (n2) = M2 × V2
\[ n_2 = 1.20 \, \text{mol/L} \times 0.250 \, \text{L} = 0.300 \, \text{mol} \]

### Step 2: Calculate the total number of moles of glucose in the final solution.

Total moles (n_total) = n1 + n2
\[ n_{\text{total}} = 0.720 \, \text{mol} + 0.300 \, \text{mol} = 1.020 \, \text{mol} \]

### Step 3: Calculate the total volume of the final solution.

Total volume (V3) = V1 + V2
\[ V_3 = 480 \, \text{mL} + 250 \, \text{mL} = 730 \, \text{mL} = 0.730 \, \text{L} \]

### Step 4: Calculate the molarity of the final solution.

Molarity (M3) = Total moles / Total volume
\[ M_3 = \frac{n_{\text{total}}}{V_3} = \frac{1.020 \, \text{mol}}{0.730 \, \text{L}} \]

Calculating this gives:
\[ M_3 \approx 1.39 \, \text{M} \]

### Final Answer:
The molarity of the final solution is approximately **1.39 M**.

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