A 1.84 mg sample of polyhydric alcoholic compound 'X' of molar mass 92.0 g/mol gave 1.344 mL of H₂ gas at STP. The number of alcoholic hydrogens present in compound 'X' is ________.

Solution

Chemistry ma’am · 4 steps

Turn both the sample and the gas into moles, then remember every H₂ carries two alcoholic hydrogens.

1. Moles of X. 1.84 mg = 1.84 × 10⁻³ g, so moles of X = (1.84 × 10⁻³)/92.0 = 2.0 × 10⁻⁵ mol.
2. Moles of hydrogen gas. At STP one mole of gas occupies 22400 mL, so moles of H₂ = 1.344/22400 = 6.0 × 10⁻⁵ mol.

   2R-OH + 2Na → 2R-ONa + H₂

3. Count the hydrogens. Each H₂ molecule comes from two O–H hydrogens, so moles of alcoholic H = 2 × 6.0 × 10⁻⁵ = 1.2 × 10⁻⁴ mol. Per molecule of X this is (1.2 × 10⁻⁴)/(2.0 × 10⁻⁵) = 6.
4. So. Compound X has 6 alcoholic hydrogens.

The common slip: 3. Stopping at the ratio of H₂ molecules to X gives 3. You have to double it, because each H₂ needs two O–H hydrogens.

The idea to keep. In active-hydrogen problems, find moles of H₂ per mole of compound, then double it. One H₂ always means two acidic hydrogens.

Answer from the NTA final answer key. Solution by PrepWiser.