Question

A metal M forms a sulphate which is isomorphous with ${\text{MgS}}{{\text{O}}_{\text{4}}}{\text{.7}}{{\text{H}}_{\text{2}}}{\text{O}}$ . If ${\text{0}}{\text{.6538 g}}$ of metal M displaced $2.16{\text{ g}}$ of silver from silver nitrate solution, then the atomic weight of the metal M is:
A. 32.61
B. 56.82
C. 65.38
D. 74.58

Answer 1

Use the concept of the number of equivalents. When one metal displaces another metal from its salt solution, the number of gram equivalents of two metals are the same.

Complete answer:
Since the metal M forms a sulphate which is isomorphous with ${\text{MgS}}{{\text{O}}_{\text{4}}}{\text{.7}}{{\text{H}}_{\text{2}}}{\text{O}}$, you can assume that the charge on the metal cation is same as the charge on the cation in ${\text{MgS}}{{\text{O}}_{\text{4}}}{\text{.7}}{{\text{H}}_{\text{2}}}{\text{O}}$ . The chemical formula ${\text{MgS}}{{\text{O}}_{\text{4}}}{\text{.7}}{{\text{H}}_{\text{2}}}{\text{O}}$ represents the salt magnesium sulphate hepta hydrate. The cation in magnesium sulphate hepta hydrate ${\text{M}}{{\text{g}}^{ + 2}}$ is magnesium cation with +2 charge.
Since you have assumed that the charge on the metal cation is same as the charge on the cation in ${\text{MgS}}{{\text{O}}_{\text{4}}}{\text{.7}}{{\text{H}}_{\text{2}}}{\text{O}}$, you can say that the charge on the metal cation is also +2.
The number of gram equivalents of the metal are equal to the number of gram equivalents of silver.
The number of gram equivalents of metals are $\dfrac{{{W_1} \times {n_1}}}{{{M_1}}}$
Here, ${W_1},{n_1},{M_1}$ represents the mass, charge and atomic weight of the metal.
The number of gram equivalents of silver are $\dfrac{{{W_2} \times {n_2}}}{{{M_2}}}$
Here, ${W_2},{n_2},{M_2}$ represents the mass, charge and atomic weight of the silver.
Since, the number of gram equivalents of the metal are equal to the number of gram equivalents of silver. You can say that $\dfrac{{{W_1} \times {n_1}}}{{{M_1}}} = \dfrac{{{W_2} \times {n_2}}}{{{M_2}}}$ … …(1).
Substitute values in the equation (1)

$$\Rightarrow \dfrac{{{W_1} \times {n_1}}}{{{M_1}}} = \dfrac{{{W_2} \times {n_2}}}{{{M_2}}} \\\ \Rightarrow \dfrac{{{\text{0}}{\text{.6538 }} \times 2}}{{{M_1}}} = \dfrac{{2.16 \times 1}}{{108}} \\\ \Rightarrow {M_1} = {\text{0}}{\text{.6538 }} \times 2 \times \dfrac{{108}}{{2.16}} \\\ \Rightarrow {M_1} = 65.38$$

Hence, the atomic weight of the metal is 65.38.

**The correct option is the option (C).

Note:**
The number of gram equivalents is the ratio of the weight to the gram equivalent weight. The gram equivalent weight is the ratio the atomic weight of the metal to its valence.