1.

When a current of 1.5 amp was passed for 30 minutes through a solution of a salt of a trivalent metal, 1.071 g of the metal was deposited at the cathode. Calculate the atomic weight of the metal.

Answer»

Solution :Charge passed through the solution `= 1.5 xx 30 xx 60`
= 2700 COULOMBS.
Mole of electric charge passed `= (2700)/(96500)` faraday `= (27)/(965)F`.
`therefore` EQUIVALENT of the metal deposited `= (27)/(965)`.
Let the at. Wt. of the metal be X.
`therefore` eq. wt. of the metal `= (x)/(3)`. `"" (because "valency of the metal = 3")`
`therefore` weight of the metal deposited = no. of eq. `xx` eq. wt.
`= (27)/(965) xx (x)/(3)g`. `therefore` as given, `(27 xx x)/(965 xx 3) = 1.071`
x = 114.8 amu.


Discussion

No Comment Found

Related InterviewSolutions