1.

A sample of solid KClO_(3) (potassium chlorate) was heated in a test tube to obtain O_(2) according to the reaction 2KClO_(3) rarr 2KCl + 3O_(2) The oxygen gas was collected by downward displacement of water at 295K. The total pressure of the mixture is 772 mm of Hg. The vapour pressure of water is 26.7 mm of Hg at 300K. What is the partial pressure of the oxygen gas ?

Answer» <html><body><p></p>Solution :`2KClO_(3(s)) <a href="https://interviewquestions.tuteehub.com/tag/rarr-1175461" style="font-weight:bold;" target="_blank" title="Click to know more about RARR">RARR</a> 2KCl_((s)) + 3O_(2 (g))` <br/> `P_("<a href="https://interviewquestions.tuteehub.com/tag/total-711110" style="font-weight:bold;" target="_blank" title="Click to know more about TOTAL">TOTAL</a>")=772mm <a href="https://interviewquestions.tuteehub.com/tag/hg-485049" style="font-weight:bold;" target="_blank" title="Click to know more about HG">HG</a>` <br/> `P_(H_(2)O) = 26.7 mm Hg` <br/> `P_("total") = P_(O_(2)) + P_(H_(2)O)` <br/> `:. P_(O_(2)) = P_("total") - P_(H_(2)O)` <br/> `P_(1) = 26.7 mm Hg T_(2) = 298K` <br/> `T_(1) = 300K, P_(2) = ?, (P_(1))/(T_(1)) = (P_(2))/(T_(2))` <br/> `P_(2) = ((P_(1))/(T_(1))) T_(2) = (26.7mm Hg)/(300 <a href="https://interviewquestions.tuteehub.com/tag/cancel-406753" style="font-weight:bold;" target="_blank" title="Click to know more about CANCEL">CANCEL</a>(K)) xx 295 cancel(K)` <br/> `P_(2) = 26.26 mm Hg` <br/> `:. P_(O_(2)) = 772-26.26` <br/> =745.74 mm Hg</body></html>


Discussion

No Comment Found

Related InterviewSolutions