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| 1. |
34.05 mL of phosphorus vapour weigh 0.0625 g at 546^(@)C and 1.0 bar pressure. What is the molar mass of phosphorus ? |
| Answer» <html><body><p><<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>></p>Solution :Step 1. Calculation of <a href="https://interviewquestions.tuteehub.com/tag/volume-728707" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUME">VOLUME</a> at `0^(@)C` and 1 <a href="https://interviewquestions.tuteehub.com/tag/bar-892478" style="font-weight:bold;" target="_blank" title="Click to know more about BAR">BAR</a> pressure <br/> `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2)), i.e., (1xx34.05)/(546+273)=(1xxV_(2))/(273)"or" V_(2)=11.35 mL` <br/> 11.35 mL of vapour at `0^(@)C` and 1 bar pressure weigh `=0.0625" <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a>"` <br/> Step 2. Calculation of mass of 22700 mL of `0^(@)C` and 1 bar pressure <br/> `:. 22700" mL"` of vapour at `0^(@)C` and 1 bar pressure will weigh `=(0.0625)/(11.35)xx22700=125" g"` <br/> `:. """Molar mass" =125" g "mol^(-1)` <br/>Alternatively, using <br/> `R=0.083" bar "<a href="https://interviewquestions.tuteehub.com/tag/dm-432223" style="font-weight:bold;" target="_blank" title="Click to know more about DM">DM</a>^(3)K^(-1)mol^(-1)` <br/> PV=nRT, i.e., `n=(PV)/(RT)=(1.0" bar"xx(34.05xx10^(-3)dm^(3))/(0.083" bar " dm^(3)K^(-1)xx819 K)=5xx10^(-4)mol` <br/> `:.`Mass of 1 mole `"=(0.0625)/(5xx10^(-4))"g"=125 g` <br/> `:. ""` Molar mass`=125 g mol^(-1)`</body></html> | |