Final answer:
To prepare a standard 0.5 molar (250 ml) aqueous solution of sodium carbonate, 13.25 g of sodium carbonate should be measured based on its molar mass of 106 g/mol. Therefore, the student who measured 13.25 g has the correct amount needed for the solution.
Step-by-step explanation:
Calculation of Correct Weight of Sodium Carbonate
To determine who measured the correct weight of sodium carbonate to prepare a standard 0.5 molar (250 ml) aqueous solution, we start by calculating the molar mass of sodium carbonate (Na2CO3). The molar mass is the sum of the atomic masses of all atoms in a molecule. For sodium carbonate:
Sodium (Na) has an atomic mass of approximately 23 u.
Carbon (C) has an atomic mass of roughly 12 u.
Oxygen (O) has an atomic mass of about 16 u.
The molecular formula of sodium carbonate Na2CO3 contains 2 sodium atoms, 1 carbon atom, and 3 oxygen atoms. Therefore, the molar mass is: (2 × 23) + 12 + (3 × 16) = 106 g/mol.
Next, we calculate the amount of sodium carbonate needed for a 0.5 M solution in 250 ml:
A 0.5 M solution contains 0.5 moles of solute per liter of solution. Since we are preparing 250 ml (0.25 L), the moles of sodium carbonate needed are (0.5 moles/L) × 0.25 L = 0.125 moles.
To find the mass required, we multiply the moles by the molar mass: 0.125 moles × 106 g/mol = 13.25 g.
Thus, to prepare a 0.5 M solution of sodium carbonate in 250 ml of water, 13.25 g of sodium carbonate should be measured. The student who measured this weight has the correct amount. Keep in mind, to ensure the accuracy of the final solution, one should always weigh the solute carefully and ensure it is dissolved completely in the solvent before making up to final volume in a volumetric flask.