Question

4 g of $NaCl$ produces 9.8 g $AgCl$ on reacting with$AgN{{O}_{3}}$. The equivalent weight of silver is (eq. wt. of$Na=23$)
(a)- 105.87
(b)- 108.57
(c)- 107.87
(d)- 102.87

Answer 1

Hint: The equivalent weight of an element is calculated by dividing the molar mass of elements with the total positive valency of the element. The number of moles is calculated by dividing the given mass to the molecular mass of the compound.

Complete answer:
The number of moles of the compound is calculated by dividing the given mass of the compound to the molar mass of the compound.
So, the number of moles of $NaCl$ = $\dfrac{\text{given mass of NaCl}}{\text{molecular mass of NaCl}}$
Given mass = 4 g
Molecular mass of $NaCl$= 23 + 35.44 = $58.44\text{ g/mol}$
So,
$\text{moles }=\text{ }\dfrac{\text{given mass of NaCl}}{\text{molecular mass of NaCl}}\text{ = }\dfrac{4}{58.44}\text{ = 0}\text{.68 moles}$
The number of moles of $AgCl$ = $\dfrac{\text{given mass of AgCl}}{\text{molecular mass of AgCl}}$
Given mass = 9.8 g
Molecular mass of $AgCl$ = 107.87 + 35.44 = $143.31\text{ g/mol}$
So, the number of moles of $AgCl$ will be,
$\text{moles }=\text{ }\dfrac{\text{given mass of AgCl}}{\text{molecular mass of AgCl}}\text{ = }\dfrac{9.8}{143.31}\text{ = 0}\text{.68 moles}$
So, it can be seen that 0.68 moles of $NaCl$ produces 0.68 moles of $AgCl$.
Therefore, the reaction will be sodium chloride and silver nitrate react to form silver chloride and sodium nitrate.

The reaction is,
$NaCl+AgN{{O}_{3}}\to AgCl+NaN{{O}_{3}}$
So, to find the equivalent weight of silver, we have to see the change in oxidation number from reactant to product.
The oxidation number of Ag in $AgN{{O}_{3}}$ = +1
The oxidation number of Ag in $AgCl$ = +1
Since, the oxidation number of Ag is the same on both sides i.e., reactant and product side.
Therefore the molecular weight of silver will be equal to the equivalent weight.
The molecular weight of silver is 107.87 g
$\text{Equivalent weight = molecular weight = 107}\text{.87}$

So, the correct answer is “Option C”.

Note: If there is any change in the oxidation number in the reaction then it must be divided by the molecular weight to get the equivalent weight. All the molecular weight must be taken exactly.