Question

A current of 0.965 ampere is passed through 500ml of 0.2m solution of znso4 for 10min.the molarity of zn2 after deposition

Answer 1

Answer: The molarity of $Zn^{2+}$ after deposition is 0.194M

Explanation:-

$Q=I\times t$

where Q= quantity of electricity in coloumbs

I = current in amperes = 0.965A

t= time in seconds =$10\times 60=600s$

$Q=0.965A\times 600s=579C$

$Zn^{2+}+2e^{-1}\rightarrow Zn$

$96500\times 2=193000Coloumb$ of electricity deposits 1 mole of  Zn.

579C of electricity deposits =$\frac{1}{193000}\times 579=0.003mole$ of  Zn.

$\text{no of moles of} Zn^{2+}={\text{Molarity}\times {\text{Volume in L}}$

$\text{no of moles of} Zn^{2+}={0.2M}\times {0.5L}=0.1mole$

Thus $Zn^{2+}$ remaining in solution = (0.1-0.003)=0.097moles

$Molarity=\frac{\text {no of moles}}{\text {Volume of solution in L}}$

$Molarity=\frac{0.097moles}{0.5L}=0.194M$

Answer 2

Answer:

_______0.194 M ______