To prepare a 0.100 mol/dm³ (or 0.100 M) solution of sodium carbonate (Na₂CO₃) in a 200 mL volumetric flask, you need to calculate the mass of sodium carbonate decahydrate (\(Na_2CO_3 \cdot 10H_2O\)) required.
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Calculate the number of moles needed:
The volume of the solution you want to prepare is 200 mL, which is equivalent to 0.200 L.
\[ \text{moles of Na₂CO₃} = \text{Concentration} \times \text{Volume} = 0.100 , \text{mol/dm}^3 \times 0.200 , \text{dm}^3 = 0.0200 , \text{mol} \]
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Calculate the molar mass of sodium carbonate decahydrate:
The molar mass of \(Na_2CO_3\):
- Sodium (Na): 23.0 g/mol × 2 = 46.0 g/mol
- Carbon (C): 12.0 g/mol = 12.0 g/mol
- Oxygen (O): 16.0 g/mol × 3 = 48.0 g/mol
So, the molar mass of \(Na_2CO_3 = 46.0 + 12.0 + 48.0 = 106.0 , \text{g/mol}\).
Now, add the mass for the water in the decahydrate:
- Water (H₂O): 18.0 g/mol × 10 = 180.0 g/mol
Therefore, the molar mass of sodium carbonate decahydrate (\(Na_2CO_3 \cdot 10H_2O\)):
\[ \text{Molar mass of } Na_2CO_3 \cdot 10H_2O = 106.0 , \text{g/mol} + 180.0 , \text{g/mol} = 286.0 , \text{g/mol} \]
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Calculate the mass of sodium carbonate decahydrate needed:
Now, use the number of moles and the molar mass to find the mass needed:
\[ \text{mass} = \text{moles} \times \text{molar mass} \]
\[ \text{mass} = 0.0200 , \text{mol} \times 286.0 , \text{g/mol} = 5.72 , \text{g} \]
So, you need to weigh off 5.72 grams of sodium carbonate decahydrate to prepare a 0.100 mol/dm³ solution in a 200 mL volumetric flask.
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