Asked by cheri
A 0.456 gram sample of an unknown mono-
protic acid (let’s call it HZ) was dissolved in
some water (you pick the amount). Then the
acidic solution was titrated to the equivalence
point with 32.5 mL of 0.174 M KOH. What
is the molecular weight of the unknown acid
HZ?
1. 103.2 g/mol
2. 187.9 g/mol
3. 80.6 g/mol
4. 85.7 g/mol
5. 0.00259 g/mol
protic acid (let’s call it HZ) was dissolved in
some water (you pick the amount). Then the
acidic solution was titrated to the equivalence
point with 32.5 mL of 0.174 M KOH. What
is the molecular weight of the unknown acid
HZ?
1. 103.2 g/mol
2. 187.9 g/mol
3. 80.6 g/mol
4. 85.7 g/mol
5. 0.00259 g/mol
Answers
Answered by
DrBob222
HZ + KOH ==> KZ + H2O
mols KOH = M x L = ?
mols HZ = mols KOH
mols = g/molar mass. You know mols and grams, solve for molar mass.
mols KOH = M x L = ?
mols HZ = mols KOH
mols = g/molar mass. You know mols and grams, solve for molar mass.