A) 1.85
B) 3.70
C) 1.085
D) 1.0425
Correct Answer: D
Solution :
\[PC{{l}_{5}}PC{{l}_{3}}+C{{l}_{2}}\] \[i=1+\alpha \] \[\alpha =\frac{\begin{align} & theoretical\text{ }vapour\text{ }density\text{ }-\text{ }experimental \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,vapour\text{ }density \\ \end{align}}{experimental\text{ }vapour\text{ }density}\] Theoretical VD, \[(D)=\frac{molar\text{ }mass}{2}\] \[\frac{208.5}{2}=104.25\] Experimental VD, \[(d)=100\] \[\because \] \[\alpha =\frac{D-d}{d}\] \[\therefore \] \[\alpha =\frac{104.25-100}{100}\] \[=\frac{4.25}{100}=0.0425\] \[i=1+\alpha =1+0.0425=1.0425\]
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