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Some Basic Concepts of Chemistry question

2023 · 31 Jan · Shift 1 · Q19

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Some Basic Concepts of Chemistry
2023 · 31 Jan · Shift 1 · Q19

Some Basic Concepts of Chemistry question

2023 · 31 Jan · Shift 1 · Q19

JEE MainChemistrySome Basic Concepts of ChemistryNumerical+4 / −1

On complete combustion,

0.492 \mathrm{~g}

of an organic compound gave

0.792 \mathrm{~g}

of

\mathrm{CO}_{2}

. The % of carbon in the organic compound is

\underline{\hspace{2cm}}

(Nearest integer)

Numerical answer

View written solutionFree

Correct answer: 44

Find the mass of carbon present in the produced $\mathrm{CO_2}$

From combustion data: $m(\mathrm{CO_2}) = 0.792\,\text{g}$

In $\mathrm{CO_2}$ ,

molar mass of $\mathrm{CO_2} = 44$
mass of carbon in 1 mole of $\mathrm{CO_2} = 12$

So, mass of carbon in $0.792\,\text{g}$ of $\mathrm{CO_2}$ is: $m(\mathrm C)=0.792\times \frac{12}{44}$

$m(\mathrm C)=0.216\,\text{g}$

Calculate percentage of carbon in the organic compound

Mass of organic compound taken: $0.492\,\text{g}$

Therefore, $\%\,\mathrm C = \frac{0.216}{0.492}\times 100$

$\%\,\mathrm C = 43.90\%$

Nearest integer

$\boxed{44}$

Comparison with stored answer

Stored correct answer = $44$

Our derived answer is also $44$ , so they agree.

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