The amount of oxalic acid required to prepare 300 mL of 2.5 M solution is : (molar mass of oxalic acid = 90 g `mol^(–1)`) :

Options

A. 67.5 g
B. 9.45 g
C. 6.75 g
D. 94.5 g

Answer

To calculate the amount of oxalic acid required to prepare 300 mL of a 2.5 M solution, we can follow these steps:

### Step 1: Understand the formula for molarity
Molarity (M) is defined as the number of moles of solute divided by the volume of the solution in liters.

\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]

### Step 2: Convert the volume from mL to L
We need to convert the volume of the solution from milliliters to liters. Since there are 1000 mL in 1 L:

\[ \text{Volume in liters} = \frac{300 \text{ mL}}{1000} = 0.3 \text{ L} \]

### Step 3: Calculate the number of moles of oxalic acid
Using the molarity formula, we can rearrange it to find the moles of solute:

\[ \text{moles of solute} = \text{Molarity} \times \text{Volume in liters} \]

Substituting the values:

\[ \text{moles of oxalic acid} = 2.5 \, \text{M} \times 0.3 \, \text{L} = 0.75 \, \text{moles} \]

### Step 4: Calculate the mass of oxalic acid required
To find the mass of oxalic acid, we use the formula:

\[ \text{mass} = \text{moles} \times \text{molar mass} \]

Given that the molar mass of oxalic acid is 90 g/mol:

\[ \text{mass of oxalic acid} = 0.75 \, \text{moles} \times 90 \, \text{g/mol} = 67.5 \, \text{g} \]

### Final Answer
The amount of oxalic acid required to prepare 300 mL of a 2.5 M solution is **67.5 g**.

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