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 hydrogen present in compound 'X' is

We have a 1.84 mg sample of a polyhydric alcohol 'X' with molar mass 92.0 g/mol, and it gives 1.344 mL of $$H_2$$ gas at STP. We need to find the number of alcoholic (-OH) hydrogens in 'X'.

We begin by finding the number of moles of compound 'X'. The mass is 1.84 mg = $$1.84 \times 10^{-3}$$ g, and the molar mass is 92.0 g/mol, so:

$$n_X = \frac{1.84 \times 10^{-3}}{92.0} = 2.0 \times 10^{-5} \text{ mol}$$

Now we find the moles of $$H_2$$ produced. At STP, 1 mole of any gas occupies 22400 mL. The volume of $$H_2$$ is 1.344 mL, so:

$$n_{H_2} = \frac{1.344}{22400} = 6.0 \times 10^{-5} \text{ mol}$$

When a polyhydric alcohol reacts with an active metal like sodium, each -OH group releases half a mole of $$H_2$$. If the compound has $$k$$ alcoholic hydrogen atoms, then one mole of the compound produces $$\frac{k}{2}$$ moles of $$H_2$$:

$$R(OH)_k + kNa \rightarrow R(ONa)_k + \frac{k}{2}H_2$$

We can write the ratio:

$$\frac{n_{H_2}}{n_X} = \frac{k}{2}$$

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

$$3 = \frac{k}{2}$$

$$k = 6$$

We can verify: a compound with molar mass 92 and 6 -OH groups would be consistent with a highly hydroxylated small molecule. Indeed, the moles and volumes check out perfectly.

Hence, the correct answer is 6.