To prepare 250 ml of m/20 oxalic acid . Using the prepared solution as a standard solution find the strength and molarity of a given potassium permagnate solution
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Hello dear,
● Answer with Explaination-
Balanced molecular eqn is-
2KMnO4 + 2H2SO4 + 5(COOH)2 ---> K2SO4 + 2MnSO4 + 18H2O + 10CO2
Let, x = volume of KMnSO4 required for titration
(a) Molarity-
M(KMnSO4)×V(KMnSO4) / M(oxalic acid)×V(oxalic acid) = 2/5
M(KMnSO4) = 1/20 × 250 × 2/5 / x
M(KMnSO4) = 5/x M
(b) Strength -
Strength(KMnSO4) = M(KMnSO4) × molecular mass
Strength(KMnSO4) = 790/x g/l
Hope this is useful...
● Answer with Explaination-
Balanced molecular eqn is-
2KMnO4 + 2H2SO4 + 5(COOH)2 ---> K2SO4 + 2MnSO4 + 18H2O + 10CO2
Let, x = volume of KMnSO4 required for titration
(a) Molarity-
M(KMnSO4)×V(KMnSO4) / M(oxalic acid)×V(oxalic acid) = 2/5
M(KMnSO4) = 1/20 × 250 × 2/5 / x
M(KMnSO4) = 5/x M
(b) Strength -
Strength(KMnSO4) = M(KMnSO4) × molecular mass
Strength(KMnSO4) = 790/x g/l
Hope this is useful...
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