Redox and Equivalent ConceptHard
Question
A quantity of 0.5 g of a metal nitrate gave 0.43 g of metal sulphate.
Options
A.The equivalent weight of the metal is 38.
B.The equivalent weight of the metal is 76.
C.The atomic weight of metal may be 76.
D.The atomic weight of metal may be 19.
Solution
neq metal nitrate = neq metal sulphate
Or $\frac{0.5}{E_{M} + 62} = \frac{0.43}{E_{M} + 48} \Rightarrow E_{M} = 38$
Create a free account to view solution
View Solution FreeMore Redox and Equivalent Concept Questions
0.116 g of C4H4O4 (A) is neutralised by 0.074 g of Ca(OH)2. Hence protonic hydrogen (H+) in (A) will be...To 1 L of 1.0 M impure H2SO4 sample, 1.0 M NaOH solution was added and a plot was obtained as follows: The % purity of H...The equivalent weight of an element is 25. If its specific heat is 0.085 cal/K-g, its exact atomic mass should be...1 g of the carbonate of a metal was dissolved in 25 mL of 1N-HCl. The resulting liquid required 5 mL of 1N-NaOH for natu...RH2 (ion exchange resin) can replace Ca2+ in hard water as follows. RH2 + Ca2+ → RCa + 2H+.One litre of hard water after...