1.00 g of a chloride of an element contains 0.835 g of chlorine. If the vapour density of the chloride is 85, find the atomic weight of the element and its valency.

Explanation

Let the formula of the chloride be `MCI_(x)` where x is the valency of the element M.
`x xx ` moles of M =1 `xx` moles of CI
or `x xx (1-0.835)/("at . Wt . Of M)= (0.835)/(35.5)`
Now we know
molecular weight `= 2xx ` vapour density
`=2xx85 =170`
For `MCI_(x)`
mol . wt. of `MCI_(x)` = at . wt. of M +x `xx` at . wt. of CI
or 170 =at .wt . of M + 35.5 x.
`:. ` at . wt. of M = 170 - 35.5 x
or `x = (170-"at .wt.of M")/(35.5)`
From (1) and (2), we get, at. wt. of M = 28.05. Substituting at. wt. of M in the eqn. (1) we get, x = 4.