KCET · Chemistry · Redox Reactions
What is the equivalent weight of \(\mathrm{SnCl}_{2}\) in the following reaction
\(\mathrm{SnCl}_{2}+\mathrm{Cl}_{2} \longrightarrow \mathrm{SnCl}_{4}\) ?
- A 95
- B 45
- C 60
- D 30
Answer & Solution
Correct Answer
(A) 95
Step-by-step Solution
Detailed explanation
\[
\begin{aligned}
\mathrm{SnCl}_{2}+\mathrm{Cl}_{2} & \longrightarrow \mathrm{SnCl}_{4} \\
190 & & \frac{190}{E_{1}} &=\frac{71}{35.5} \\
\Rightarrow & E_{1} &=95
\end{aligned}
\]
\begin{aligned}
\mathrm{SnCl}_{2}+\mathrm{Cl}_{2} & \longrightarrow \mathrm{SnCl}_{4} \\
190 & & \frac{190}{E_{1}} &=\frac{71}{35.5} \\
\Rightarrow & E_{1} &=95
\end{aligned}
\]
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