For the dissolution of `1.08 g` of metal, `0.49 g` of `H_(2)SO_(4)` was required. If specific heat of metal is `0.06 cal//g`, what is its atomic mass?

Explanation

App. At. Wt. `= (6.4)/(0.06) = 10.6.67`
Let Eq. wt. of metal be `E`.
`(1.08)/(E) = (0.49)/(49)`
`:. E_("metal") = 108`
Also valence `= ("at.wt.")/("Eq. wt.") = (106.67)/(108) = 1` (integer)
`:.` Exact at wt. `= "Eq. wt" xx " valence" = 108 xx 1 = 108`