Question

Pure benzoic acid is not available. The benzoic standard must be prepared from sodium benzoate, calculate the mass of sodium benzoate required to prepare 100.0 mL of 100 ppm benzoic acid

Answer 1

100 ppm = 100 mg/L

It means, 100 mg is dissolved in 1 L

So, the concentration of benzoic acid in 1 L is = 100 * 10^-3/122.12 = 8.188 *10^-4 moles

             Concentration of benzoic acid in 100 ml is = 100 * 10^-3/122.12 = 8.188 *10^-5 moles

As we know that one mole of sodium benzoate will produce one mole of benzoic acid

So, 8.188 *10^-5 moles will give 8.188 *10^-5 moles of benzoic acid.

Molecular weight of sodium benzoate is 144.11 g/mol

8.188 *10^-5 moles = Weight / 144.11 g/mol

8.188 *10^-5 moles *144.11 g/mol = Weight

1.179*10^-2 g = Weight

The mass of sodium benzoate required to prepare 100.0 mL of 100 ppm benzoic acid is 1.179*10^-2 g or 0.01179 g.