86.3k views
1 vote
"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."

User Mehul
by
5.5k points

1 Answer

4 votes

Answer:

60.44 g

Step-by-step 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

User Jeemusu
by
5.8k points