A 1.84 mg sample of polyhydric alcoholic compound 'X' of molar mass 92.0 g/mol gave 1.344 mL of H_2 gas at STP. The number of alcoholic hydrogens present in compound 'X' is \underline\hspace2cm .

Some Basic Concepts of Chemistry PYQ — JEE Main 2022 · 29 Jul · Shift 2 · Q21 with Solution

View written solutionFree

Correct answer: 6

A polyhydric alcohol reacts with sodium as:

\mathrm{R(OH)}_n + n\,\mathrm{Na} \rightarrow \mathrm{R(ONa)}_n + \frac{n}{2}\,\mathrm{H_2}

So, if a molecule has $n$ alcoholic hydrogens, then 1 mole of the compound gives:

\frac{n}{2} \text{ mol of } \mathrm{H_2}

Given mass of sample:

1.84\,\text{mg} = 1.84 \times 10^{-3}\,\text{g}

Molar mass of $X$ :

92.0\,\text{g mol}^{-1}

Hence moles of $X$ :

\text{moles of } X = \frac{1.84 \times 10^{-3}}{92.0} = 2.0 \times 10^{-5}\,\text{mol}

Volume of $\mathrm{H_2}$ at STP:

1.344\,\text{mL} = 1.344 \times 10^{-3}\,\text{L}

At STP, molar volume is:

22.4\,\text{L mol}^{-1}

So moles of $\mathrm{H_2}$ are:

\text{moles of } \mathrm{H_2} = \frac{1.344 \times 10^{-3}}{22.4} = 6.0 \times 10^{-5}\,\text{mol}

If each molecule has $n$ alcoholic hydrogens, then:

2.0 \times 10^{-5} \times \frac{n}{2} = 6.0 \times 10^{-5}

Simplifying:

1.0 \times 10^{-5} \cdot n = 6.0 \times 10^{-5}

n = 6

The number of alcoholic hydrogens in compound $X$ is:

\boxed{6}

Stored correct answer = $6$ .

My derived answer also is $6$ , so they agree.