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 find the amount of oxalic acid required to prepare a 300 mL solution of 2.5 M concentration, 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. The formula is:
\[ M = \frac{n}{V} \]
where:
- \( M \) = molarity (in moles per liter, mol/L)
- \( n \) = number of moles of solute
- \( V \) = volume of the solution (in liters)

### Step 2: Convert volume from mL to L
Given that the volume of the solution is 300 mL, we need to convert this to liters:
\[ V = \frac{300 \text{ mL}}{1000} = 0.3 \text{ L} \]

### Step 3: Calculate the number of moles of oxalic acid
Using the molarity formula, we can rearrange it to find the number of moles:
\[ n = M \times V \]
Substituting the values:
\[ n = 2.5 \, \text{mol/L} \times 0.3 \, \text{L} = 0.75 \, \text{mol} \]

### Step 4: Calculate the mass of oxalic acid required
To find the mass of oxalic acid, we use the formula:
\[ \text{mass} = n \times \text{molar mass} \]
Given that the molar mass of oxalic acid is 90 g/mol:
\[ \text{mass} = 0.75 \, \text{mol} \times 90 \, \text{g/mol} = 67.5 \, \text{g} \]

### Conclusion
The amount of oxalic acid required to prepare 300 mL of a 2.5 M solution is **67.5 grams**.

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