Question

A metal M forms a sulphate which is isomorphous with mgso4.7h2o .If 0.638 g of metal M displaced 2.16 g of silver from silver nitrate sol. Then atomic weight of metal is

Answer 1

let metal 1 and silver 2

w1/E1=w2/E2
0.638/E1=2.16/108
E1=32.69
atomic weight of metal would be =32.69×2 (V.F.)
=65.38

Hope it will help you.

Answer 2

Answer : The atomic weight of metal is, 63.8 grams

Explanation:

According to law of chemical Equivalence,

Gram equivalent of metal = Gram equivalent of silver

$\frac{w_1\times n_1}{M_1}=\frac{w_2\times n_2}{M_2}$

where,

$w_1$ = mass of metal = 0.638 g

$w_2$ = mass of silver = 2.16 g

$n_1$ = charge on cation = 2

$n_2$ = charge on silver = 1

$M_1$ = atomic mass of metal = ?

$M_2$ = atomic mass of silver = 108 g/mole

Now put all the given values in the above formula, we get the atomic mass of metal.

$\frac{0.638\times 2}{M_1}=\frac{2.16\times 1}{108}$

$M_1=63.8g$

Therefore, the atomic weight of metal is, 63.8 grams