QJEEM23Ch756c: JEE Main 2023 Chemistry - Class 11 JEEM23Ch756c Solution: An organic compound gives of and of on complete combustion. If…

To find the percentage of hydrogen in the compound, we first need to determine the ratio of moles of carbon and hydrogen in the products.

The moles of CO₂ produced can be calculated by dividing the mass of CO₂ produced by its molar mass (44.01 g/mol):

$\text{moles of CO}_{2} = \frac{0.220 \,\mathrm{g}}{44.01 \,\mathrm{g/mol}} = 0.005 \, \mathrm{mol}$

Since each mole of CO₂ contains one mole of carbon, there are 0.005 moles of carbon in the compound.

Now, let's calculate the moles of H₂O produced:

$\text{moles of H}_{2}\text{O} = \frac{0.126 \,\mathrm{g}}{18.02 \,\mathrm{g/mol}} = 0.007 \, \mathrm{mol}$

Since each mole of H₂O contains two moles of hydrogen, there are 0.014 moles of hydrogen in the compound.

Now, let's find the masses of carbon and hydrogen in the compound:

Mass of carbon = moles of carbon × molar mass of carbon

$\text{Mass of carbon} = 0.005 \, \mathrm{mol} \times 12.01 \,\mathrm{g/mol} = 0.060 \,\mathrm{g}$

Mass of hydrogen = moles of hydrogen × molar mass of hydrogen

$\text{Mass of hydrogen} = 0.014 \, \mathrm{mol} \times 1.008 \,\mathrm{g/mol} = 0.014 \,\mathrm{g}$

Now, we are given that the percentage of carbon is 24%. Let's find the total mass of the compound:

$\text{Total mass of compound} = \frac{\text{Mass of carbon}}{\% \text{ of carbon}} = \frac{0.060 \,\mathrm{g}}{0.24} = 0.250 \,\mathrm{g}$

Finally, let's find the percentage of hydrogen:

$\% \text{ of hydrogen} = \frac{\text{Mass of hydrogen}}{\text{Total mass of compound}} \times 100 = \frac{0.014 \,\mathrm{g}}{0.250 \,\mathrm{g}} \times 100 = 5.6$

So, the percentage of hydrogen is 5.6% or 56 × 10^-1.