Prepare a 1.00 L of a 0.25 M solution of sodium carbonate. Determine the amount of sodium carbonate needed in correct significant figures. A)21 g B)26 g C)332 g D)424 g?

To prepare a 1.00 L solution of 0.25 M sodium carbonate (Na₂CO₃), you can use the formula:

[ \text{mass (g)} = \text{molarity (M)} \times \text{volume (L)} \times \text{molar mass (g/mol)} ]

The molar mass of Na₂CO₃ is approximately 106 g/mol. Thus, the calculation is:

[ 0.25 , \text{M} \times 1.00 , \text{L} \times 106 , \text{g/mol} = 26.5 , \text{g} ]

Rounding to two significant figures gives 26 g, so the correct answer is B) 26 g.