To find out how many grams of benzoic acid are needed to make a 521.0 mL solution at a concentration of 0.47 m, we can use the formula:
grams of solute = molarity (M) × volume (L) × molar mass (g/mol)
First, we need to convert the volume from mL to liters:
521.0 mL = 521.0 / 1000 = 0.521 L
Now we can plug in the values:
- Molality (M) = 0.47 m
- Volume (L) = 0.521 L
- Molar Mass of benzoic acid = 122.13 g/mol
Now, calculating:
grams of benzoic acid = 0.47 mol/L × 0.521 L × 122.13 g/mol
grams of benzoic acid ≈ 28.32 g
Rounding this to two significant figures (based on the molality provided, 0.47), we find:
grams of benzoic acid ≈ 28 g
Thus, you would need approximately 28 g of benzoic acid to make 521.0 mL of a 0.47 m solution.