(a) 0.456 g of a metal gives 0.606 g of its chloride. Calculate the equivalent mass of the metal
(b)Calculate the equivalent mass of potassium dichromate. The reduction half-reaction in acid medium is,
`Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-) rarr 2Cr^(3+)+7H_(2)O`

Explanation

Mass of the metal = `W_(1)` = 0.456 g.
Mass of the metal chloride = `W_(2)` = 0.606 g
`:.` Mass of chlorine = `W_(2) -W_(1)` = 0.606 - 0.456 = 0.15 g
0.15 g of chlorine combine with 0.456 g of metal.
`:.` 35.46 g of chlorine will combine with`0.456/0.15 xx 35.46` = 107.79 g `eq ^(-1)`.
`Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-) rarr 2Cr^(3+)+7H_(2)O`
Equivalent mass = `("Molar mass")/("No. of electrons gained")`
`:.` Molar mass of `K_(2)Cr_(2)O_(7)` = 294.18
`:.` Equivalent mass of `K_(2)Cr_(2)O_(7) = 294.18/6 = 49.03`