Question

A Zn rod weighing $25$g was kept in $100$ mL of $1$M $CuS{O}_4$ solution. After a certain time the molarity of $C{u}^{2+}$ in solution was $0.8$. What was the weight of Zn rod after cleaning? At. weight of Zn $=65.4$

• A. $23.69$gm
• B. $45.2$gm
• C. $32$gm
• D. $54$gm

Answer 1

The correct option is A $23.69$gm

Weight of zinc rod = 25 g

Number of moles that are required by $C{u}^{2+}toCu=Molarity\times Volume=(1-0.8)\times \frac{100L}{1000}=0.2\times 0.1=0.02mole$

Molar mass of zinc = $65.4gmo{l}^{-1}$

The zinc that gets oxidized = $0.02\times 65.4$ g = 1.308 g

Hence, weight of Zn rod left will be = 25 - 1.308 = 23.69 gm