Chlorine reacts with NaOH: cold dilute, Cl2 + 2NaOH -> NaCl + NaOCl + H2O; hot concentrated, 3Cl2 + 6NaOH -> 5NaCl + NaClO3 + 3H2O. For 0.3 mol of Cl2 in each case, select the correct statement(s).

Hot reaction: 0.5 mol of electrons are transferred

Cold reaction: 0.3 mol of NaOCl is formed

Cold reaction: n-factor of Cl2 is 1

Hot reaction: 0.5 mol of electrons are transferred

Cold reaction: 0.3 mol of NaOCl is formed

Hot reaction: 0.3 mol of NaClO3 is formed

Cold reaction: n-factor of Cl2 is 1

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3rx165

Chlorine reacts with $NaOH$ : cold dilute, $Cl X_{2} + 2 NaOH NaCl + NaOCl + H X_{2} O$ ; hot concentrated, $3 Cl X_{2} + 6 NaOH 5 NaCl + NaClO X_{3} + 3 H X_{2} O$ . For $0.3$ mol of $Cl X_{2}$ in each case, select the correct statement(s).

Hot: 5 electrons per 3

Cl X_{2}

, so

0.3 \times 5/3 = 0.5

mol e

^{-}

(A true);

NaClO X_{3} = 0.1

mol (C false). Cold: 1

NaOCl

per

Cl X_{2}

, so 0.3 mol (B true); one electron is lost per

Cl X_{2}

, so

n = 1

(D true).