A solution of a salt of a metal was electrolysed for 150 minutes by passing 0.15 A current. The weight of the metal deposited was 0.783 g. The specific heat of the metal is `0.057" cal"//gK`. The atomic mass X of the metal is :A. `111.80" g"//mol`B. `52.2" g"//mol`C. `200" g"//mol`D. `250" g"//mol`

A solution of a salt of a metal was electrolysed for 150 minutes by pa

1.	A solution of a salt of a metal was electrolysed for 150 minutes by passing 0.15 A current. The weight of the metal deposited was 0.783 g. The specific heat of the metal is `0.057" cal"//gK`. The atomic mass X of the metal is :A. `111.80" g"//mol`B. `52.2" g"//mol`C. `200" g"//mol`D. `250" g"//mol`
Answer» Correct Answer - A (a) According to Dulong & Petits Law, Approximate Atomic mass`=(6.4" Cal"//K)/(0.057"Cal"//gK)=112.28" g"` Quantity of charge passed `=(0.5" amp")xx(150xx60 s)` `=1350" amp"-s=1350" C"` 1350 C charge deposit metal =0.738 g 96500 C charge deposit metal `=((0.738g))/((1350c))xx(96500c)=55.9 g` `:.` Equivalent mass of metal=55.97 g Approximate valency`=("App.Atomic mass")/("Equivalent mass")` `=((112.28g))/((55.97 g))=2.006` Since valency is a whole number, Exact valency =2 Exact atomic mass`="Eq.mass"xx"valency"` `55.97xx2=111.94 g`

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A solution of a salt of a metal was electrolysed for 150 minutes by passing 0.15 A current. The weight of the metal deposited was 0.783 g. The specific heat of the metal is `0.057" cal"//gK`. The atomic mass X of the metal is :A. `111.80" g"//mol`B. `52.2" g"//mol`C. `200" g"//mol`D. `250" g"//mol`

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