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If 7.15mL of 0.101M NaOH solution is required to neutralize 0.178g of an unknown acid, what...

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?

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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

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