A) \[Xe{{F}_{2}}\]
B) \[Xe{{F}_{6}}\]
C) \[Xe{{F}_{4}}\]
D) None of these
Correct Answer: B
Solution :
| \[\frac{{{r}_{1}}}{{{r}_{2}}}=\frac{{{P}_{1}}}{{{P}_{2}}}\sqrt{\frac{{{M}_{2}}}{{{M}_{1}}}}\] |
| \[\frac{\left( \frac{{{n}_{1}}}{{{t}_{1}}} \right)}{\left( \frac{{{n}_{2}}}{{{t}_{2}}} \right)}=\frac{{{P}_{1}}}{{{P}_{2}}}\sqrt{\frac{{{M}_{2}}}{{{M}_{1}}}}\,\,\,\,\left( r=\frac{n}{t} \right)\] |
| \[\frac{\frac{1}{38}}{\left( \frac{1}{57} \right)}=\frac{0.8}{1.6}\sqrt{\frac{{{M}_{2}}}{28}}\] \[\Rightarrow {{M}_{2}}=252\,\,gm\] |
| so formula of compound is \[Xe{{F}_{6}}\] |
| \[\because \,\,\,(131+6\times 19=245)\] |
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