Question

71 g of chlorine combines with a metal giving 111g of its chloride. The chloride is isomorphous with $MgC{l_2}.6{H_2}O$ . The atomic mass of the metal is:
(A) 20
(B) 30
(C) 40
(D) 69

Answer 1

In crystallography crystals are described as isomorphous which is the similarity in shape of crystals. Crystal shape is defined by measuring angles between crystal faces. We calculate the weight of the metal and since it is isomorphous, we can find its equivalent weight and multiply it with its valency to get its atomic weight.
${E_M} = \dfrac{{{W_M} \times {M_{C{l_2}}}}}{{{W_{C{l_2}}}}}$

Complete Step by step solution:
Here, chlorine combines with a certain metal and forms a certain amount of chloride compound of the metal. The weight of chlorine which combines with the metal is 71 g. The amount of chloride compound formed due to the reaction of chlorine and metal is 111 g. Hence the amount of metal in the chloride compound of 111 g, will be 40 g.
Weight of $C{l_2}$ = ${W_{C{l_2}}}$ = 71 gram
Weight of metal = ${W_M}$ = $111-71 = 40$ gram
As the metal chloride formed is isomorphous with $MgC{l_2}.6{H_2}O$ , both will have similar chemical formulas. So, the atoms of both the compounds will have the same valency. Hence the valency of the metal will be the same with that of a magnesium atom. The valency of the metal will be 2.
Equivalent weight of metal is the gram atomic weight of an element divided by its valency. So, to find the atomic weight of an element, we multiply the equivalent weight of the metal with its valency. Let ${E_M}$ be the equivalent weight of the metal.
And hence, ${E_M} = \dfrac{{{W_M} \times 35.5}}{{{W_{C{l_2}}}}} = \dfrac{{40 \times 35.5}}{{71}} = 20$
${A_M} = {E_M} \times 2 = 20 \times 2 = 40$ amu
Therefore, the atomic mass of the metal is 40 amu.
The correct answer is option C.

Note:
The law of isomorphism is given by the German chemist Eilhard Mistscherich. Isomorphous compounds have similar chemical formulas. The element forming isomorphous compounds must have the same valency.