1.

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.

Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/let-11597" style="font-weight:bold;" target="_blank" title="Click to know more about LET">LET</a> the formula of the chloride be `MCI_(<a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a>)` where x is the valency of the <a href="https://interviewquestions.tuteehub.com/tag/element-969236" style="font-weight:bold;" target="_blank" title="Click to know more about ELEMENT">ELEMENT</a> M. <br/> `x <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> ` moles of M =1 `xx` moles of CI <br/> or `x xx (1-0.835)/("at . Wt . Of M)= (0.835)/(35.5)` <br/> Now we know <br/> molecular weight `= 2xx ` vapour density <br/> `=2xx85 =170`<br/> For `MCI_(x)` <br/> mol . wt. of `MCI_(x)` = at . wt. of M +x `xx` at . wt. of CI <br/> or 170 =at .wt . of M + 35.5 x. <br/> `:. ` at . wt. of M = 170 - 35.5 x <br/>or `x = (170-"at .wt.of M")/(35.5)` <br/> From (1) and (2), we get, at. wt. of M = 28.05. Substituting at. wt. of M in the eqn. (1) we get, x = <a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>.</body></html>


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