Answer:
(c) \[\underset{1}{\mathop{\,_{17}^{37}C}}\,\underset{:}{\mathop{:}}\,\,{{\underset{3}{\mathop{_{17}^{35}Cl}}\,}_{Ratio}}\]
Average atomic mass \[=\frac{(1\times
37)+(3\times 35)}{1+3}\]
\[=\frac{142}{4}=35.5\]
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