3.575grams
From the question, we will need to determine the moles of Na₂CO3. 10H₂O using the formula below;
[tex]moles=molarity\times volume[/tex]Given the following parameters
Molarity = 0.05M
Volume = 250ml = 0.25L
Determine the moles of Na₂CO3. 10H₂O
[tex]\begin{gathered} moles=0.05\times0.25 \\ moles=0.0125moles \end{gathered}[/tex]Determine the Na₂CO3. 10H₂O
[tex]\begin{gathered} Mass=moles\times molar\text{ mass} \\ Mass=0.0125moles\times286\frac{g}{mol} \\ mass=3.575grams \end{gathered}[/tex]Hence the mass of Na₂CO3. 10H₂O required to prepare a 250 ml of a 0.05 M solution is 3.575grams
