$2.10 g$ of the carbonate $MCO X_{3}$ of an unknown divalent metal $M$ is exactly neutralised by $51.2 mL$ of $0.800 N$ $HCl$ . Taking the equivalent mass of the carbonate ion as $30$ , find the atomic mass of $M$ (two decimals).

Numerical answer type, answer in g/mol, tolerance 0.05.

Answer: 42.54 g/mol

Equivalents of

HCl = 0.800 \times 0.0512 = 0.04096

, so equivalent mass of

MCO X_{3} = 2.10/0.04096 = 51.27

. Equivalent mass of the metal

=

that of the salt

-

that of

CO X_{3} X^{2 -} = 51.27 - 30 = 21.27

. For a divalent metal, atomic mass

= 2 \times 21.27 = 42.54