Question

$4g$ of $NaCl$ produce $9.8g$ AgCl on reacting with ${AgNO}_3$. The equivalent wt. of silver is (Eq. wt of Na =23):

• A. $105.87$
• B. $108.57$
• C. $107.87$
• D. $102.87$

Answer 1

The correct option is C $107.87$

Solution:- (C) 107.87

As we know that,

$\text{no. of moles}=\frac{\text{wt.}}{\text{mol. wt.}}$

weight of $NaCl=4g$

Molecular weight of $NaCl=58.44g/mol$

No. of moles of $AgCl=\frac{4}{58.44}=0.68\text{moles}$

weight of $NaCl=9.8g$

Molecular weight of $AgCl=143.32g/mol$

No. of moles of $AgCl=\frac{9.8}{143.32}=0.68\text{moles}$

$NaCl+AgN{O}_3⟶AgCl+NaN{O}_3$

Oxidation number of $Ag$ in $AgN{O}_3$ is $+1$.

Oxidation number of $Ag$ in $AgCl$ is $+1$.

As the oxidation number of $Ag$ in both, i.e., reactant and product side, is same.

$\therefore$ Equivalent weight of $Ag=$ Molecular weight $=107.8682g\approx 107.87g$

Hence the required answer is $\left(C\right)107.87$.