A current of 1.5A is passed …

Step I: Quantity of electricity Q= I x t

I=1.5 A, t=1 hr = 60 x 60 s =3600 s

Current efficiency 90 %, Q = (90/100) x 1.5 x 3600 s = 4860 C

Zn²⁺ + 2e^- ---------------> Zn

Therefore, by applying Faraday law of electrolysis

2F= 2 x 96500 C of electricity deposits = 1 mol of Zinc

4860 C deposits                                 =  4860/(2 x 96500) mol of Zinc= 0.0252 mol of Zinc

Step 2: Molarity = no.of moles of solute/ volume of solution in L

0.25 = no.of moles of solute/500 x 10^-3

Therefore no. of moles of solute = 0.25 x 500 x 10^-3 = 0.125 moles

Therefore no. of moles Zn⁺ left = (0.125-0.0252) moles= 0.0998 moles

Hence req. molarity = 0.0998 /(500 x 10^-3) = 0.1996 M

Alternatively

equivalent of .of Zn²⁺ lost  = 4860/96500=0.0504 eq

Therefore meq of Zn²⁺ lost = 0.0504 x 10³=50.3

Initial meq of Zn²⁺ = 500 x 0.25x 2= 250 meq (500 mL of 0.25 M solution)

Hence meq of Zn²⁺ remains in the solution = (250 - 50.3) meq =199.7 meq

Therefore molarity of the remaining solution = 199.7/(2 x500) M= 0.1997 M