A) \[{{P}_{2}}\]
B) \[{{P}_{4}}\]
C) \[{{P}_{3}}\]
D) \[{{P}_{8}}\]
Correct Answer: B
Solution :
Vapour density of phosphorus at \[310{}^\circ C\] and 775 mm Hg is 2.64 g/l. i.e., \[{{P}_{1}}=775\,mm=\frac{775}{760}atm\] \[{{T}_{1}}=310+273=583\,K\] \[d=\frac{w}{V}=2.64\,\text{g/l}\] According to gas-equation: \[PV=nRT=\frac{w}{m}RT\] or \[m=\frac{w}{V}\cdot \frac{RT}{P}\] or \[m=\frac{2.64\times 0.0821\times 583}{775\text{/}760}\] Atomic weight of phosphorus is 31. Hence number of atoms bonded together to form one molecule will be\[=\frac{124}{31}=4.\] Hence, molecular formula of phosphorus\[={{P}_{4}}.\]
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