A 1.84 mg sample of polyhydric alcoholic compound 'X' of molar mass 92.0 g/mol gave 1.344 mL of $\mathrm{H}_{2}$ gas at STP. The number of alcoholic hydrogens present in compound 'X' is ________.

<p>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:</p> <p>$n = \frac{V}{V_m}$ </p> <p>where:</p> <ul> <li>$n$ is the number of moles,</li> <li>$V$ is the volume of the gas, and</li> <li>$V_m$ is the molar volume of the gas.</li> </ul> <p>Substituting in the given values:</p> <p>$n = \frac{1.344 \text{ mL}}{22400 \text{ mL/mol}} = 5.995 \times 10^{-5} \text{ mol}$</p> <p>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:</p> <p>$n_H = 2 \times n = 2 \times 5.995 \times 10^{-5} \text{ mol} = 1.199 \times 10^{-4} \text{ mol}$</p> <p>This number of moles of hydrogen represents the number of moles of alcoholic hydrogen atoms in the $1.84$ mg sample of the compound &#39;X&#39;. </p> <p>We can determine the number of moles of compound &#39;X&#39; in the sample by dividing the mass of the sample by the molar mass of the compound:</p> <p>$n_X = \frac{m}{M}$</p> <p>where:</p> <ul> <li>$n_X$ is the number of moles of &#39;X&#39;,</li> <li>$m$ is the mass of &#39;X&#39;, and</li> <li>$M$ is the molar mass of &#39;X&#39;.</li> </ul> <p>Substituting in the given values:</p> <p>$n_X = \frac{1.84 \text{ mg}}{92 \text{ g/mol}} = \frac{1.84 \times 10^{-3} \text{ g}}{92 \text{ g/mol}} = 2.0 \times 10^{-5} \text{ mol}$</p> <p>The number of alcoholic hydrogens per molecule of &#39;X&#39; is then given by the ratio of the number of moles of hydrogen to the number of moles of &#39;X&#39;:</p> <p>$\frac{n_H}{n_X} = \frac{1.199 \times 10^{-4} \text{ mol}}{2.0 \times 10^{-5} \text{ mol}} = 6$</p> <p>So, compound &#39;X&#39; contains $6$ alcoholic hydrogens.</p>