A. 0.085 mols/dm^3 x 98 g/mol = ? grams/dm^3
B. volume H2SO44 = 27.30 cc. volume KOH = 25.00 cc
Titration to the M.O. end point means both Hs of the H2SO4 were titrated
H2SO4 + 2KOH ==> K2SO4 + 2H2O
mols H2SO4 = M x L = 0.085 x 0.02730 = about 0.0023 but that a close estimate and you should redo it more accurately
mols KOH = 2 x mols H2SO4 = about 0.0046
M KOH = mols KOH/L KOH = about 0.0046/0.02500 = about 0.18 Mol/dm^3
C. KOH = about 0.18 mol/dm^3
0.18 mols/dm^3 x molar mass KOH = g/dm^3
Solution A in H2So4 containing
0.085mol /dm^3. solution B is
KoH. Titrate using methyl orange
as an indicator. calculate:
A.concentration of A in g/dm^3
B.Concentration of B In mol/dm^3
C.Concentration of B in g/dm^3
take Va=27.30cm^3,Vb=25.0
0cm^3?
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