1.00 g of a mixture, consisting of equal number of moles of carbonates of two alkali metals, required 44.4 " mL of " 0.5 N-HCl for complete reaction. If the atomic weight of one of the metal is 7.00. Find the atomic weight of the other metal. What will be the total amount of sulphate formed on quantitative conversion of 1.00 g of the mixture into sulphates?

Explanation

Total `HCl=(44.4xx0.5)/(1000)=0.0222eq`
`M_(2)CO_(3)+2"HCl"to2MCl+H_(2)O+CO_(2)`
`2 " Eq of "HCl=1 " mol of "CO_(3)^(2-)`
`0.0222 " Eq of "HCl=(0.0222)/(2)=0.011 mol of` total `CO_(3)^(2-)`
Since moles are equal, therefore, the number of moles of `M_(2)CO_(3)`, is equal to the number of moles of `M_(2)'CO_(3)` i.e., `(0.0111)/(2)` moles of each metal carbonate required.
Molecular weight of `M_(2)CO_(3)=2xx7+60=74`
Molecular weight of `M_(2)'CO_(3)=(2x+60)`
Weight of `M_(2)CO_(3)=(74xx0.0111)/(2)`
Weight of `M_(2)'CO_(3)=(2x+60)xx(0.0111)/(2)`
`therefore(74xx0.0111)/(2)+((2x+60)xx0.0111)/(2)=1.0g` (given)
`thereforex=23.09`
Atomic weight of the other metal`=23.09`
For the formation of sulphates:
`M_(2)CO_(3)+H_(2)SO_(4)toM_(2)SO_(4)+H_(2)O+CO_(2)`
`(0.0111)/(2)` " mol of "`M_(2)CO_(3)=(0.0111)/(2)` " mol of "sulphate
Total weight of sulphates
`(0.0111xx110)/(2)+(0.0111xx142.18)/(2)=1.4g`