Question

A current of 0.965 ampere is passed through 500 ml of 0.2 M solution of $ZnS{O}_4$ for 10 minutes. The molarity of $Z{n}^{2+}$ after deposition of zinc is:

• A. 0.1 M
• B. 0.5 M
• C. 0.8 M
• D. 0.194 M

Answer 1

The correct option is D 0.194 M

$Z{n}^{2+}+2e^{-}⟶Zn$

$1$ mole is deposited by $2\times 96500=193000C$

The charge supplied = $0.965\times 10\times 60=579C$

$\therefore$ Number of moles deposited = $\frac{579}{193000}=0.003$ moles

In $500ml$, number of moles = $\frac{500}{1000}\times 0.2=0.1$ moles

$\therefore$ Moles remaining in the $500ml$ solution = $0.1-0.003=0.097$ moles

$\therefore$ Molarity = $\frac{1000}{500}\times 0.097=0.194M$

Option D is correct.