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34.05 mL of phosphorus vapour weights 0.0625 g at 546^(@)C and 0.1 bar pressure. What is the molar mass of phosphorus ?

Answer» <html><body><p></p>Solution :According to ideal <a href="https://interviewquestions.tuteehub.com/tag/gas-1003521" style="font-weight:bold;" target="_blank" title="Click to know more about GAS">GAS</a> law,<br/>`pV=nRT=(w)/(M)RT`<br/>`therefore M=(wRT)/(pV)`<br/>where,<br/>Molecular <a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> of phosphorus, (w) = 0.0625 g<br/> Gas Constant `(R )=8.314xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)` <a href="https://interviewquestions.tuteehub.com/tag/bar-892478" style="font-weight:bold;" target="_blank" title="Click to know more about BAR">BAR</a> `L mol^(-1)K^(-1)` <br/> Pressure of phosphorus vapour (p) = 1.0 bar<br/>Absolute <a href="https://interviewquestions.tuteehub.com/tag/temperature-11887" style="font-weight:bold;" target="_blank" title="Click to know more about TEMPERATURE">TEMPERATURE</a> `(T)=(546+273)K=819 K`<br/>Volume of Phosphorus vapour (V) = 34.05 mL<br/>= 0.03405 L<br/>`therefore M =((0.0625 g)(8.314xx10^(-2)"bar L mol"^(-1)K^(-1))(819K))/(("1 bar")(0.03405 L))`<br/>`= 124.98 = 125 g mol^(-1)`<br/>Note : If 0.1 bar is corrct for according to text book then answer will be 1249.8. It is not real because,<br/>`P_(4)=31xx4=124`</body></html>


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