A) \[{{103}^{o}}\]
B) \[{{107}^{o}}\]
C) \[{{109}^{o}}28\]
D) \[\underline{X}\]
Correct Answer: C
Solution :
% of Cd in \[CdC{{l}_{2}}=\frac{0.9}{1.5}\times 100\] \[=60%\] Therefore, % of \[C{{l}_{2}}\] in \[CdC{{l}_{2}}=100-60=40%\] \[\because \] 40% part \[(C{{l}_{2}})\] has atomic weight \[=2\times 35.5=71.0\] \[\therefore \]60% part (Cd) has atomic weight \[=\frac{71.0\times 60}{40}\] \[=106.5\]
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