Given:
Current (I) = 1.4 A
Volume of solution (V) = 500 mL = 0.5 L
Concentration of zinc sulfate (M) = 0.180 M
Time (t) = 200 seconds

To find:
Molarity of Zn2+ ions after deposition of zinc

Explanation:
Zinc sulfate (ZnSO4) dissociates in water to form Zn2+ ions and SO4^2- ions.

The current passing through the solution causes the Zn2+ ions to gain electrons and get reduced to form zinc (Zn) metal at the cathode. At the same time, the SO4^2- ions move towards the anode.

Step 1: Calculate the number of moles of ZnSO4:
Molarity (M) = moles/volume (L)
moles = M x volume
moles of ZnSO4 = 0.180 M x 0.5 L = 0.090 moles

Step 2: Calculate the charge passed through the solution:
Charge (Q) = current (I) x time (t)
Q = 1.4 A x 200 s = 280 C

Step 3: Calculate the number of moles of electrons:
1 mole of electrons = 1 Faraday = 96485 C
moles of electrons = Q/F
moles of electrons = 280 C / 96485 C/mol = 0.0029 moles

Step 4: Calculate the number of moles of Zn2+ ions reduced:
The balanced chemical equation for the reduction of Zn2+ ions is:
Zn2+ + 2e- -> Zn

From the equation, it is clear that 1 mole of Zn2+ ions is reduced by 2 moles of electrons.
moles of Zn2+ ions reduced = 0.0029 moles x (1 mole Zn2+/2 moles e-) = 0.00145 moles

Step 5: Calculate the new molarity of Zn2+ ions:
New moles of Zn2+ ions = initial moles of Zn2+ ions - moles of Zn2+ ions reduced
new moles of Zn2+ ions = 0.090 moles - 0.00145 moles = 0.08855 moles

new molarity of Zn2+ ions = new moles of Zn2+ ions / volume (L)
new molarity of Zn2+ ions = 0.08855 moles / 0.5 L = 0.1771 M

Therefore, the molarity of Zn2+ ions after the deposition of zinc is 0.1771 M.