Answer:
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.
