Set - I	Set - II
A) $\%$ of 'S' in ${\text{SO}}_2$	1) 27.27
B) $\%$ of 'C' in ${\text{CO}}_2$	2) 80
C) $\%$ of 'C' in Ethane	3) 40
D) $\%$ of 'C' in Glucose	4) 50

For A: The gram molecular weight of  ${\text{SO}}_2$ is 64g                                                                                                                                                                                   

The gram atomic weight of S is 32g

            $\text{S\hspace{0.17em}=\hspace{0.17em}32}$ ; ${\text{O}}_{\text{2}}\text{=\hspace{0.17em}32}$; ${\text{SO}}_{\text{2}}=64$

${\text{\%of S in\hspace{0.17em}SO}}_2=\frac{\text{Weight\hspace{0.17em}of\hspace{0.17em}sulphur}}{{\text{Gram\hspace{0.17em}molecular\hspace{0.17em}weight\hspace{0.17em}of\hspace{0.17em}SO}}_{\text{2}}}\times 100$

             Therefore,   $\%S=\frac{32}{64}\times 100=50\%$

For B:  The gram molecular weight of ${\text{CO}}_{\text{2}}$$\text{=\hspace{0.17em}44}$

Molar mass of carbon,C$\text{=\hspace{0.17em}}12$

             $\%\text{C}=\frac{12}{44}\times 100=27.27$

For C: The molecular formula of ethane is ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$.

  Then total mass of C in ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$, $\text{2\hspace{0.17em}C =2}\times \text{12 =24}$ C=24,

Molar mass of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ $\text{=\hspace{0.17em}30}$

            Now,  $\%\text{C}=\frac{24}{30}\times 100=80\%$

For D: The chemical formula of glucose is ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$.

 Total mass of carbon ,C $=12\times 6=72$

Molar mass of glucose, ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$$=180$

            Then, $\%\text{C}=\frac{72}{180}\times 100=40\%$