7.38 g of a sample of a metal oxide is quantitatively reduced to 6.84 g of pure metal. If the specific heat of the metal is 0.0332 cal/g, calculate the valency and the accurate atomic weight of the metal.

Explanation

Let the formula of the oxide be `M_(2)O_(x)` , where x is the valency of the metal M. We have, therefore,
`x xx `moles of M = `2 xx` moles of O.
We know that, atomic weight `xx` specific heat `~~` 6.4
(Dulong and Petit.s law)
`:.` approximate atomic weight `= (6.4)/(0.0332) = 193 `
From eqn. `(1), x xx (6.64)/(193) =2xx ((7.38-6.84))/(16)`
or
x= 1.9.
But valency is always a whole number and so the valency of the metal is 2. Now, to calculate the accurate value of the atomic weight of the metal, substitute the value of x as 2 in equation (1) again,
or `2xx(6.84)/("atomic wt.") = 2xx ((7.38 - 6.84))/(16)`
Atomic weight (accurate) = 202.67.