To calculate the amount of anhydrous sodium carbonate ([tex]\( \text{Na}_2\text{CO}_3 \)[/tex]) required to prepare 100 mL of a 0.5 Normal (N) solution, we will follow these steps:
1. Determine the molecular weight of [tex]\( \text{Na}_2\text{CO}_3 \)[/tex]:
- Sodium (Na) has an atomic mass of approximately 23 g/mol.
- Carbon (C) has an atomic mass of approximately 12 g/mol.
- Oxygen (O) has an atomic mass of approximately 16 g/mol.
- Therefore, the molecular weight of [tex]\( \text{Na}_2\text{CO}_3 \)[/tex] is calculated as:
[tex]\[
2 \times 23 \, (\text{Na}) + 12 \, (\text{C}) + 3 \times 16 \, (\text{O}) = 46 + 12 + 48 = 106 \, \text{g/mol}
\][/tex]
2. Determine the equivalent weight of [tex]\( \text{Na}_2\text{CO}_3 \)[/tex]:
- Since [tex]\( \text{Na}_2\text{CO}_3 \)[/tex] provides 2 equivalents of Na ions per molecule, the equivalent weight is the molecular weight divided by 2:
[tex]\[
\text{Equivalent weight of } \text{Na}_2\text{CO}_3 = \frac{106 \, \text{g/mol}}{2} = 53 \, \text{g/equiv}
\][/tex]
3. Convert the desired volume from milliliters to liters:
- Given volume is [tex]\( 100 \, \text{mL} \)[/tex]:
[tex]\[
\text{Volume in liters} = \frac{100 \, \text{mL}}{1000} = 0.1 \, \text{L}
\][/tex]
4. Calculate the required amount of [tex]\( \text{Na}_2\text{CO}_3 \)[/tex]:
- Using the formula:
[tex]\[
\text{grams} = \text{Normality} \times \text{Equivalent weight} \times \text{Volume} (in \, L)
\][/tex]
- Substitute the values:
[tex]\[
\text{grams of } \text{Na}_2\text{CO}_3 = 0.5 \, \text{N} \times 53 \, \text{g/equiv} \times 0.1 \, \text{L} = 2.65 \, \text{g}
\][/tex]
Thus, to prepare 100 mL of a 0.5 N solution of anhydrous sodium carbonate, you will need 2.65 grams of [tex]\( \text{Na}_2\text{CO}_3 \)[/tex].
