Redox and Equivalent ConceptHard
Question
x g of KHC2O4 requires 100 ml of 0.02 M-KMnO4 in acidic medium. In another experiment, y g of KHC2O4 requires 100 ml of 0.05 M-Ca(OH)2. The ratio of x and y is
Options
A.1: 1
B.1: 2
C.2: 1
D.5: 4
Solution
$n_{eq}MnO_{4}^{-} = n_{eq}FeSO_{4}$
Or $\frac{x}{M} \times 2 = \frac{100 \times 0.02}{1000} \times 5(1)$
$n_{eq}KHC_{2}O_{4} = n_{eq}Ca(OH)_{2} $$${\text{Or }\frac{y}{M} \times 1 = \frac{100 \times 0.5}{1000} \times 2(2) }{\therefore\frac{x}{y} = \frac{1}{2}}$$
Create a free account to view solution
View Solution FreeMore Redox and Equivalent Concept Questions
The oxidation state of C in C6H12O6 is equal to the oxidation state of C in...A volume of 100 L of hard water requires 5.6 g of lime for removing temporary hardness. The temporary hardness in ppm of...To an acidic solution of an anion, a few drops of KMnO4 solution are added. Which of the following , if present, will no...A certain amount of a reducing agent reduces x mole of KMnO4 and y mole of K2Cr2O7 in different experiments in acidic me...A small amount of CaCO3 completely neutralizes 525 ml of 0.1 N-HCl and no acid is left at the end. After converting all ...