21.3 g of Cl2 is passed through excess hot concentrated NaOH (3Cl2 + 6NaOH -> 5NaCl + NaClO3 + 3H2O). Molar masses (g mol^-1): Cl2 71, NaClO3 106.5, NaCl 58.5. The mass of NaClO3 formed is

JEE Advanced levelSingle correct optionDifficulty 3 of 3ph196

$21.3 g$ of $Cl X_{2}$ is passed through excess hot concentrated NaOH ( $3 Cl X_{2} + 6 NaOH 5 NaCl + NaClO X_{3} + 3 H X_{2} O$ ). Molar masses ( $g mo l^{- 1}$ ): $Cl X_{2}$ 71, NaClO3 106.5, NaCl 58.5. The mass of NaClO3 formed is

n (Cl X_{2}) = 21.3/71 = 0.300 mol

. Three mol of

Cl X_{2}

give 1 mol of

NaClO X_{3}

, so

n (NaClO X_{3}) = 0.300/3 = 0.100 mol

. Mass

= 0.100 \times 106.5 = 10.65 g

. The other options correspond to 0.300 mol of NaOCl (cold-alkali product), 0.500 mol of NaCl and 0.300 mol of NaClO3.