Question

If 7.15mL of 0.101M NaOH solution is required to neutralize 0.178g of an unknown acid, what is the molecular weight of the unknown acid?

Answer 1

first calcolate mole of NaOH

no. of mole = molarity volumle of solution in liter

7.15 ml = 0.00715 L

no. of mole of NaOH = 0.101 0.00715 = 0.00072215 mole of NaOH

so to nutrilize acid 0.00072215 mole of NaOH thus mole of acid = 0.00072215 mole

0.00072215 mole = 0.178 gm then 1 mole of acid = 10.178/0.00072215 = 246.48 gm/mol

molecular weight of unknown acid is 246.48 gm