$21.3 g$ of $Cl X_{2}$ is passed into hot concentrated NaOH so that all of it disproportionates to chloride and chlorate. Taking $Na = 23$ and $Cl = 35.5$ , the mass of $NaCl$ formed is

Answer: A

3 Cl X_{2} + 6 NaOH 5 NaCl + NaClO X_{3} + 3 H X_{2} O

. Moles of

Cl X_{2} = 21.3/71 = 0.30

, so

NaCl = 0.30 \times 5/3 = 0.50

mol. Mass

= 0.50 \times 58.5 = 29.25 g

. (The value 17.55 g would be the cold, dilute NaOH result, where 1 mol NaCl forms per mol

Cl X_{2}