To determine the weight of anhydrous sodium trioxocarbonate (IV) (Na₂CO₃) required to prepare a 0.05 M solution in a 250 mL volumetric flask, follow these steps:
1. Identify the given quantities:
- Molarity (M): [tex]\( 0.05 \, \text{M} \)[/tex]
- Volume (V): [tex]\( 250 \, \text{mL} \)[/tex] which is [tex]\( 0.250 \, \text{L} \)[/tex] (since 1000 mL = 1 L)
- Molar mass (Mₘ) of Na₂CO₃: [tex]\( 105.99 \, \text{g/mol} \)[/tex]
2. Calculate the number of moles of Na₂CO₃ required:
To calculate the moles required, use the equation:
[tex]\[
\text{Moles} = \text{Molarity} \times \text{Volume}
\][/tex]
Substituting the given values:
[tex]\[
\text{Moles} = 0.05 \, \text{M} \times 0.250 \, \text{L} = 0.0125 \, \text{moles}
\][/tex]
3. Calculate the weight (mass) of Na₂CO₃ needed:
To find the mass, use the equation:
[tex]\[
\text{Mass} = \text{Moles} \times \text{Molar mass}
\][/tex]
Substituting the values:
[tex]\[
\text{Mass} = 0.0125 \, \text{moles} \times 105.99 \, \text{g/mol} = 1.324875 \, \text{g}
\][/tex]
Therefore, the weight of anhydrous sodium trioxocarbonate (IV) required to prepare exactly 0.05 M of its solution in a 250 mL volumetric flask is 1.324875 grams.
