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  • The complete combustion of 0.492 g of an organic compound containing 'C', 'H' and 'O' gives 0.793g of <img alt="\mathrm{CO}_{2} \text { and } 0.442 \mathrm{~g} \te

The complete combustion of 0.492 g of an organic compound containing 'C', 'H' and 'O' gives 0.793g of \mathrm{CO}_{2} \text { and } 0.442 \mathrm{~g} \text { of } \mathrm{H}_{2} \mathrm{O}. The percentage of oxygen composition in the organic compound is __________.(nearest integer)

Option: 1

46


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Given, Organic compound =0.492 \mathrm{~g}

           \begin{aligned} &\mathrm{CO}_{2}=0.793 \mathrm{~g} . \\\\ &\mathrm{H_{2} }\mathrm{O}=0.442 \mathrm{~g} . \end{aligned}

\text { moles of } \mathrm{CO}_{2}=\text { moles of } \mathrm{C} \text { in product formed }

                               =\mathrm{\frac{0.793}{44}}

\mathrm{\therefore \text { Weight of } C=\frac{0.793}{ 44} \times 12=0.216 \mathrm{~g}}
                      

\text { moles of 'H' }=\frac{0.442}{18} \times 2

\text { Weight of 'H'}=\frac{0.442}{18} \times 2 \times 1=0.049 \mathrm{~g}\\

\therefore \text { Weight of } \mathrm{O}= \text { Total wt. - wt.of } \mathrm{H}-\mathrm{w t \cdot of \: \mathrm{C}}\\

                              \mathrm{=0.492-0.216-0.049=0.227 \; g }

\% \; \text{O}=\frac{0.227}{0.492} \times 100=46.13\; \% .

Hence, 46 is correct answer

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