The equivalence point is reached after adding 20.77 mL of base. The molar mass of unknown acid is 60 g/mol
given that :
mass of the monoprotic acid = 0.125 g
molarity of NaOH = 0.1003 M
volume = 20.77 mL = 0.02077 L
the number of moles = molarity × volume in L
= 0.1003 × 0.02077
= 0.00208 mol
moles = mass / molar mass
molar mass = mass / moles
= 0.125 g / 0.00208 mol
= 60 g /mol
Thus, A 0.125-gram sample of a monoprotic acid of unknown molar mass is dissolved in water and titrated with 0.1003 M NaOH. The equivalence point is reached after adding 20.77 mL of base. the molar mass of the unknown acid is 60 g/mol.
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