Question

"A 0.2720 g sample of the acid HA, was titrated with standard sodium hydroxide, NaOH, solution. Determine the molecular weight of the acid if the sample required 45.00 mL of 0.1000 M NaOH to neutralize."

Answer 1

Answer:

60.44 g

Explanation:

45.00 mL x 0.1000 M NaOH x     1 mol HA    

                    1000 mL NaOH      1 mol NaOH

= 0.0045 mol HA

 0.2720 g HA  

0.0045 mol HA

= 60.44 g/mol