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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.

Answer» <html><body><p></p>Solution :Let the formula of the oxide be `M_(2)O_(x)` , where x is the valency of the <a href="https://interviewquestions.tuteehub.com/tag/metal-1094457" style="font-weight:bold;" target="_blank" title="Click to know more about METAL">METAL</a> M. We have, therefore, <br/> `x xx `moles of M = `2 xx` moles of O.<br/>We know that, atomic weight `xx` specific <a href="https://interviewquestions.tuteehub.com/tag/heat-21102" style="font-weight:bold;" target="_blank" title="Click to know more about HEAT">HEAT</a> `~~` 6.4 <br/> (Dulong and Petit.s law) <br/> `:.` approximate atomic weight `= (6.4)/(0.0332) = 193 ` <br/> From <a href="https://interviewquestions.tuteehub.com/tag/eqn-973463" style="font-weight:bold;" target="_blank" title="Click to know more about EQN">EQN</a>. `(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>), x xx (6.64)/(193) =2xx ((7.38-6.84))/(16)`<br/> or <br/>x= 1.9. <br/> 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, <br/> or `2xx(6.84)/("atomic <a href="https://interviewquestions.tuteehub.com/tag/wt-1462289" style="font-weight:bold;" target="_blank" title="Click to know more about WT">WT</a>.") = 2xx((7.38 - 6.84))/(16)`<br/>Atomic weight (accurate) = 202.67.</body></html>


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