When `4.215 g` of a metallic carbonate was heated in a hard glass tube, the `CO_(2)` evolved was found to measure `1336 mL` at `27^(@)C` and `700 mm` pressure. What is the equivalent weight of the metal?

Explanation

Volume of `CO_2` at NTP `= ( 1336 xx 273)/(300) xx (700)/( 760)`
`= 1120` mL.
Suppose the equivalent weight of the metal is E.
`therefore` equivalent weight of the metal carbonate `= (E + 30)`
`(because "eq. wt. of"CO_(3)^(2-) = ( 60)/( 2) = 30)`.
Now, equivalent of metallic carbonate = equivalent of `CO_2`
`(4.215)/( E+30) = ( 1120("vol. at NTP") )/( 11200 ("vol. of 1 eq. at NTP") )`
`therefore E=12.15`.