A solution of an unknown acid is prepared by dissolving 5.573 g of the solid acid in sufficient DI water to make 300.00 mL of solution 19.92 mL of a 0.1253 M NaOH solution are required to neutralize 10.00 mL of this acid solution? What is the equivalent molar mass of the acid?
moles of NaOH required to neutralize 10.00 mL of the acid = MxV
= 0.1253 mol/L x 19.92 mL x (1L / 1000 mL) = 0.00250 mol
Hence moles of acid in 10.00 mL of the acid solution = moles of NaOH = 0.00250 mol acid
Hence moles of acid in 300.00 mL of the acid solution = (0.00250 mol acid / 10 mL solution) x 300 mL solution
= 0.075 mol
=> total moles of acid = 0.075 mol
=> mass of acid / molar mass = 0.075 mol
=> 5.573 g / molar mass = 0.075 mol
=> molar mass = 5.573 g / 0.075 mol = 74.3 g/mol (answer)
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