Question

$C{{l}_{2}}$ gas is passed aqueous $NaOH$ solution under given conditions:
(A)- $C{{l}_{2}}+NaOH(cold)\to$
(B)- $C{{l}_{2}}+NaOH(hot)\to$
Equivalent mass of $C{{l}_{2}}$ in (A) and (B) are:
(a)- Equivalent mass of $C{{l}_{2}}$ in (A) is 71.0 g / equivalent and Equivalent mass of $C{{l}_{2}}$ in (B) is 42.6 g / equivalent.
(b)- Equivalent mass of $C{{l}_{2}}$ in (A) is 142.0 g / equivalent and Equivalent mass of $C{{l}_{2}}$ in (B) is 85.2 g / equivalent.
(c)- Equivalent mass of $C{{l}_{2}}$ in (A) is 142.0 g / equivalent and Equivalent mass of $C{{l}_{2}}$ in (B) is 42.6 g / equivalent.
(d)- Equivalent mass of $C{{l}_{2}}$ in (A) is 71.0 g / equivalent and Equivalent mass of $C{{l}_{2}}$ in (B) is 85.2 g / equivalent.

Answer 1

When the chlorine is reacted with a cold sodium hydroxide solution then there is the formation of sodium chloride, sodium hypochlorite, and water. When the chlorine is reacted with hot sodium hydroxide solution then there is a formation of sodium chloride, sodium chlorate, and water. The equivalent weight or mass of the compound can be calculated by multiplying the molecular mass with the number of molecules and then dividing it with n-factor (change in oxidation number from reactant to product).

Complete step by step answer:
So there are two conditions given in which the chlorine gas is reacted with an aqueous solution of sodium hydroxide.
So in (A), the chlorine is reacted with a cold solution of sodium hydroxide, so the products will be sodium chloride, sodium hypochlorite, and water. The balanced chemical reaction is given below:
$C{{l}_{2}}+2NaOH\to NaCl+NaOCl+{{H}_{2}}O$
So the oxidation number of chlorine in the reactant is zero and there are two products in which chlorine atom is present i.e., sodium chloride in which the oxidation number of chlorine is -1 and sodium hypochlorite in which the oxidation number of chlorine is +1. So for both the products the change in the number of electrons is 1. So the n-factor is 1.
The equivalent mass of chlorine can be calculated by multiplying the molecular mass with the number of molecules and then dividing it with n-factor. The number of moles in the reactant is 1.
So we can write:
$Equivalent\text{ }mass=\dfrac{molar\text{ }mass\text{ }of\text{ }chlorine\text{ x }1}{n-factor}$
The molar mass of chlorine is 70.0 g / mol (35. + 35.5)
So the eq. mass will be = $=\dfrac{\text{70}\text{.0 x }1}{1}=70.0$
So the equivalent mass is 70.0 g / equivalent.
So in (B), the chlorine is reacted with a cold solution of sodium hydroxide, so the products will be sodium chloride, sodium chlorate, and water. The balanced chemical reaction is given below:
$3C{{l}_{2}}+5 NaOH\to 5 NaCl+NaCl{{O}_{3}}+3{{H}_{2}}O$
So the oxidation number of chlorine in the reactant is zero and there are two products in which chlorine atom is present i.e., sodium chloride in which the oxidation number of chlorine is -1 and in sodium chlorate, the oxidation number of chlorine is +5. So for both the products the change in the number of electrons is 5. So the n-factor is 5.
The equivalent mass of chlorine can be calculated by multiplying the molecular mass with the number of molecules and then dividing it with n-factor. The number of moles in the reactant is 3.
So we can write:
$Equivalent\text{ }mass=\dfrac{molar\text{ }mass\text{ }of\text{ }chlorine\text{ x 3}}{n-factor}$
The molar mass of chlorine is 70.0 g / mol (35. + 35.5)
So the eq. mass will be = $=\dfrac{\text{70}\text{.0 x 3}}{5}=42.6$
So the equivalent mass is 42.6 g / equivalent.

Therefore, the correct answer is option (a).

Note: It must be noted that the reaction must be balanced otherwise the number of moles of the reactant will be wrong. The change in the oxidation state should be taken the highest number of two products are given.