Question

34.05 mL of phosphorus vapour weighs 0.0625 g at 546°C and 0.1 bar pressure. What is the molar mass of phosphorus ?

Answer 1

From given,

Initial pressure, ${ P }_{ 1 }$ = 0.1

Volume, ${ V }_{ 1 }$ = 34.05 ml

Volume, ${ V }_{ 2 }$ = ?

Final temperature, ${ T }_{ 2 }$ = 546°C

Initial temperature, ${ T }_{ 2 }$ = 273°C

Volume ${ 0 }^{ o }C$ and 1 bar of pressure =$\quad \frac { { P }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } \quad =\quad \frac { { P }_{ 2 }{ V }_{ 2 } }{ { T }_{ 2 } } \quad$  

$\frac { 0.1\quad \times \quad 34.05 }{ 546\quad +\quad 273 } \quad =\quad \frac { 1\quad \times \quad { V }_{ 2 } }{ 273 } \\\\ { V }_{ 2 }\quad =\quad \frac { 0.1\quad \times \quad 34.05\quad \times \quad 273 }{ (546\quad +\quad 273) } \quad =\quad 1.135\quad ml$

Weight of 1.135 ml of vapour at ${ 0 }^{ 0 }C$ and 1 bar of pressure = 0.065 g  

$Weight\quad of\quad 22700\quad ml\quad of\quad vapour\quad at\quad { 0 }^{ 0 }C\quad and\quad 1\quad barpressure,\\\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { 0.0625 }{ 1.135 } \times \quad 22700\quad =\quad 1250\quad g$

Molar mass, M = 1250 g