Question

The volatile chloride of an element has a vapour density `approx69`. One gram of the chloride on hydrolysis yields hydrochloric acid and compound free of chlorine. Addition hydrochloric acid and compound free of chlorine. Addition of `AgNO_(3)` to this solution precipitates 3.129 g of AgCl. What may be the atomic weight of the element.?

Answer 1

Molecular weight`=2xx69=138`. Since 1 g of thwe chloride gives 3.129 g of AgCl, it follows that 1 mol, i.e., 138 g, would yield `(138xx3.129)/(143.5)` moles of `AgClapprox3` " mol of "AgCl. Formula of the chloride `=XCl_(3)`
Therefore, atomic weight of `X=(138-3xx35.5)=31.5`