What amount of bromine will be required to convert 2 g of phenol into 2,4,6-tribromophenol? (Given molar mass in $$gmol^{-1}$$ of C, H, O, Br are 12, 1, 16, 80 respectively)

The bromination of phenol to form 2,4,6-tribromophenol follows the reaction: $$ C_6H_5OH + 3Br_2 \to C_6H_2Br_3OH + 3HBr $$. For 2 g of phenol, the molar mass is calculated as $$(C_6H_5OH) = 6(12) + 6(1) + 16 = 94$$ g/mol, while bromine has a molar mass of $$Br_2 = 2 \times 80 = 160$$ g/mol.

The number of moles of phenol is $$ n_{phenol} = \frac{2}{94} \text{ mol} $$. From stoichiometry, 1 mole of phenol requires 3 moles of bromine, so $$ n_{Br_2} = 3 \times \frac{2}{94} = \frac{6}{94} = \frac{3}{47} \text{ mol} $$. The mass of bromine required is $$ m_{Br_2} = \frac{3}{47} \times 160 = \frac{480}{47} = 10.21 \text{ g} \approx 10.22 \text{ g} $$. Therefore, the correct answer is Option 4: 10.22 g.