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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? |
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Answer» 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` |
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