You used 1.512 g of your unknown weak acid and it took 37.98 mL of your standardized 0.107 M NaOH to reach the 2nd equivalence point, using the procedure described in the lab manual. What is the molecular weight of the acid

Explanation:

37.98 mL of .107 M NaOH = 37.98 x .107 mL of M NaOH

= 4.064 mL  of M NaOH .

= 4.064 x 10⁻³ L of M NaOH

It will contain 4.064 x 10⁻³ moles of NaOH

4.064 x 10⁻³ moles of NaOH will react with 4.064 x 10⁻³ moles of unknown acid

1.512 g of acid = 1.512 / M moles of acid where M is molecular weight of acid

1.512 / M = 4.064 x 10⁻³

M = 1.512 /  4.064 x 10⁻³

= 372 .

Mol weight of acid = 372.