QJEEM22Chfdcc: JEE Main 2022 Chemistry - Class 11 JEEM22Chfdcc Solution: A 1.84 mg sample of polyhydric alcoholic compound 'X' of molar…

STP conditions define the volume of 1 mole of any gas as $22.4$ liters or $22400$ mL. The volume of hydrogen gas evolved here is $1.344$ mL, so we can calculate the number of moles of hydrogen gas ( $H_2$ ) using the formula:

$n = \frac{V}{V_m}$

where:

• $n$ is the number of moles,
• $V$ is the volume of the gas, and
• $V_m$ is the molar volume of the gas.

Substituting in the given values:

$n = \frac{1.344 \text{ mL}}{22400 \text{ mL/mol}} = 5.995 \times 10^{-5} \text{ mol}$

Each molecule of $H_2$ contains $2$ atoms of hydrogen. Therefore, the number of moles of hydrogen atoms is twice the number of moles of hydrogen gas:

$n_H = 2 \times n = 2 \times 5.995 \times 10^{-5} \text{ mol} = 1.199 \times 10^{-4} \text{ mol}$

This number of moles of hydrogen represents the number of moles of alcoholic hydrogen atoms in the $1.84$ mg sample of the compound 'X'.

We can determine the number of moles of compound 'X' in the sample by dividing the mass of the sample by the molar mass of the compound:

$n_X = \frac{m}{M}$

where:

• $n_X$ is the number of moles of 'X',
• $m$ is the mass of 'X', and
• $M$ is the molar mass of 'X'.

Substituting in the given values:

$n_X = \frac{1.84 \text{ mg}}{92 \text{ g/mol}} = \frac{1.84 \times 10^{-3} \text{ g}}{92 \text{ g/mol}} = 2.0 \times 10^{-5} \text{ mol}$

The number of alcoholic hydrogens per molecule of 'X' is then given by the ratio of the number of moles of hydrogen to the number of moles of 'X':

$\frac{n_H}{n_X} = \frac{1.199 \times 10^{-4} \text{ mol}}{2.0 \times 10^{-5} \text{ mol}} = 6$

So, compound 'X' contains $6$ alcoholic hydrogens.