Answer :
To determine the amount of hydrated oxalic acid required to prepare 500 mL of a 0.1 N (Normal) solution, follow these steps:
1. Understand the terms and given values:
- Volume of the solution: 500 mL
- Normality (N): 0.1
- Molar mass of hydrated oxalic acid [tex]\((\text{H}_2\text{C}_2\text{O}_4 \cdot 2\text{H}_2\text{O})\)[/tex]: 126.07 g/mol
2. Convert the volume to liters:
[tex]\[ \text{Volume in liters} = \frac{500 \text{ mL}}{1000} = 0.5 \text{ L} \][/tex]
3. Calculate the equivalents of oxalic acid needed:
Normality (N) is defined as the number of equivalents of solute per liter of solution. Therefore:
[tex]\[ \text{Equivalents of oxalic acid} = \text{Normality} \times \text{Volume in liters} = 0.1 \times 0.5 = 0.05 \text{ equivalents} \][/tex]
4. Identify the relationship between equivalents and moles for oxalic acid:
Hydrated oxalic acid ([tex]\(\text{H}_2\text{C}_2\text{O}_4 \cdot 2\text{H}_2\text{O}\)[/tex]) is diprotic, meaning it can donate 2 hydrogen ions (H⁺) per molecule. Therefore, 1 mole of oxalic acid provides 2 equivalents.
Thus, the number of moles of hydrated oxalic acid is:
[tex]\[ \text{Moles of oxalic acid} = \frac{\text{Equivalents of oxalic acid}}{2} = \frac{0.05}{2} = 0.025 \text{ moles} \][/tex]
5. Calculate the mass of hydrated oxalic acid required:
The mass can be calculated using the molar mass:
[tex]\[ \text{Mass of oxalic acid} = \text{Moles of oxalic acid} \times \text{Molar mass} = 0.025 \text{ moles} \times 126.07 \text{ g/mol} = 3.15175 \text{ grams} \][/tex]
Therefore, the amount of hydrated oxalic acid required to prepare 500 mL of a 0.1 N solution is 3.15175 grams.
1. Understand the terms and given values:
- Volume of the solution: 500 mL
- Normality (N): 0.1
- Molar mass of hydrated oxalic acid [tex]\((\text{H}_2\text{C}_2\text{O}_4 \cdot 2\text{H}_2\text{O})\)[/tex]: 126.07 g/mol
2. Convert the volume to liters:
[tex]\[ \text{Volume in liters} = \frac{500 \text{ mL}}{1000} = 0.5 \text{ L} \][/tex]
3. Calculate the equivalents of oxalic acid needed:
Normality (N) is defined as the number of equivalents of solute per liter of solution. Therefore:
[tex]\[ \text{Equivalents of oxalic acid} = \text{Normality} \times \text{Volume in liters} = 0.1 \times 0.5 = 0.05 \text{ equivalents} \][/tex]
4. Identify the relationship between equivalents and moles for oxalic acid:
Hydrated oxalic acid ([tex]\(\text{H}_2\text{C}_2\text{O}_4 \cdot 2\text{H}_2\text{O}\)[/tex]) is diprotic, meaning it can donate 2 hydrogen ions (H⁺) per molecule. Therefore, 1 mole of oxalic acid provides 2 equivalents.
Thus, the number of moles of hydrated oxalic acid is:
[tex]\[ \text{Moles of oxalic acid} = \frac{\text{Equivalents of oxalic acid}}{2} = \frac{0.05}{2} = 0.025 \text{ moles} \][/tex]
5. Calculate the mass of hydrated oxalic acid required:
The mass can be calculated using the molar mass:
[tex]\[ \text{Mass of oxalic acid} = \text{Moles of oxalic acid} \times \text{Molar mass} = 0.025 \text{ moles} \times 126.07 \text{ g/mol} = 3.15175 \text{ grams} \][/tex]
Therefore, the amount of hydrated oxalic acid required to prepare 500 mL of a 0.1 N solution is 3.15175 grams.
