5.3% (w/v) Na₂CO₃ solution and 6.3% (w/v) H₂C₂O₄ 2H₂O solution have same

To solve this problem, we need to understand the concept of weight/volume percentage (w/v%) and molarity. The formula to calculate molarity is:

Molarity (M) = (Weight of Solute in grams / Molar Mass of Solute) / Volume of Solution in Liters

For the two solutions given:

• Na₂CO₃ solution: We have a 5.3% (w/v) solution. This means 5.3 grams of Na₂CO₃ are dissolved in 100 mL of solution.
• H₂C₂O₄ 2H₂O solution: We have a 6.3% (w/v) solution. This means 6.3 grams of H₂C₂O₄ 2H₂O are dissolved in 100 mL of solution.

Now, we will calculate the molarity of each solution.

Step-by-Step Calculation:

1. Na₂CO₃ Molarity Calculation:
   Molar mass of Na₂CO₃ = 105.99 g/mol. 
   Molarity = (5.3 g / 105.99 g/mol) / 0.1 L = 0.5 M.
2. H₂C₂O₄ 2H₂O Molarity Calculation:
   Molar mass of H₂C₂O₄ 2H₂O = 126.07 g/mol.
   Molarity = (6.3 g / 126.07 g/mol) / 0.1 L = 0.5 M.

Conclusion: Both solutions have the same molarity, which is 0.5 M.