To calculate the amount of anhydrous sodium carbonate ([tex]\(Na_2CO_3\)[/tex]) required to prepare 100 mL of a semi-normal solution, let's follow these steps:
### Step 1: Calculate the Molecular Weight of [tex]\(Na_2CO_3\)[/tex]
The molecular weight of [tex]\(Na_2CO_3\)[/tex] is determined by adding the atomic weights of its constituent atoms:
- Sodium (Na) has an atomic weight of 23.
- Carbon (C) has an atomic weight of 12.
- Oxygen (O) has an atomic weight of 16.
Given the molecular formula [tex]\(Na_2CO_3\)[/tex]:
- There are 2 sodium atoms, so the combined weight from sodium is [tex]\(2 \times 23 = 46\)[/tex].
- There is 1 carbon atom, so the weight from carbon is [tex]\(1 \times 12 = 12\)[/tex].
- There are 3 oxygen atoms, so the combined weight from oxygen is [tex]\(3 \times 16 = 48\)[/tex].
Adding these together:
[tex]\[
\text{Molecular weight of } Na_2CO_3 = 46 + 12 + 48 = 106 \, \text{g/mol}
\][/tex]
### Step 2: Determine the Number of Milliequivalents Required
- We are asked to prepare 100 mL (which is 0.1 L) of a semi-normal solution (0.5N).
Normality (N) is given, and volume (V) is converted into liters:
[tex]\[
\text{Normality (N)} = 0.5 \, \text{N}
\][/tex]
[tex]\[
\text{Volume (V)} = 0.1 \, \text{L}
\][/tex]
The number of milliequivalents (meq) is calculated as:
[tex]\[
\text{meq} = \text{N} \times \text{V} = 0.5 \times 0.1 = 0.05 \, \text{eq}
\][/tex]
### Step 3: Calculate the Number of Moles Required
For [tex]\(Na_2CO_3\)[/tex], each mole provides 2 equivalents (since 1 mole of [tex]\(Na_2CO_3\)[/tex] dissociates to give 2 moles of [tex]\(Na^+\)[/tex]).
Therefore, the number of moles required (moles_required) is:
[tex]\[
\text{moles_required} = \frac{\text{meq}}{\text{equivalents per mole}} = \frac{0.05}{2} = 0.025 \, \text{moles}
\][/tex]
### Step 4: Determine the Mass of [tex]\(Na_2CO_3\)[/tex] Required
Finally, we calculate the mass of [tex]\(Na_2CO_3\)[/tex] required using its molecular weight. The mass (mass_Na2CO3_required) is:
[tex]\[
\text{mass_Na2CO3_required} = \text{moles_required} \times \text{molecular weight of } Na_2CO_3 = 0.025 \times 106 = 2.65 \, \text{g}
\][/tex]
Thus, to prepare 100 mL of a semi-normal [tex]\( Na_2CO_3 \)[/tex] solution, you need 2.65 grams of anhydrous sodium carbonate.
