Two voltameters, one containing a solution of silver salt and the...

According to Faraday's first law of electrolysis, the mass of the substance deposited is given by:

$m = \dfrac{M}{nF} I t$

where $M$ is the atomic weight, $n$ is the valency, $F$ is Faraday's constant ($96500 \text{ C mol}^{-1}$), $I$ is the current, and $t$ is the time in seconds.

Given:
$m = 1.00 \text{ g}$
$n = 3$ (for a trivalent metal)
$I = 2 \text{ A}$
$t = 1.50 \text{ hours} = 1.50 \times 3600 \text{ s} = 5400 \text{ s}$

Substituting the values into the formula:

$1.00 = \dfrac{M \times 2 \times 5400}{3 \times 96500}$

$M = \dfrac{1.00 \times 3 \times 96500}{2 \times 5400}$

$M = \dfrac{289500}{10800}$

$M = 26.8 \text{ g mol}^{-1}$