MHTCET 15 May 2024 Shift 1 Paper

Concept:Use Faraday's laws of electrolysis to relate charge passed to the mass of calcium deposited at the cathode.Explanation:First, convert time to seconds.$t = 60 \times 60 = 3600 \, \text{s}$Calculate the total charge $Q$ passed.$Q = I \times t = 0.8 \, \text{A} \times 3600 \, \text{s} = 2880 \, \text{C}$Calcium forms $\mathrm{Ca}^{2+}$ ions, so $n = 2$ electrons are involved per ion.Use the electrolysis formula:$$m = \frac{Q \times M}{n \times F}$$Substitute the known values: $Q = 2880 \, \text{C}$, $M = 40 \, \text{g mol}^{-1}$, $n = 2$, and $F = 96500 \, \text{C mol}^{-1}$.$$m = \frac{2880 \times 40}{2 \times 96500}$$Simplify the expression.$$m = \frac{115200}{193000} \approx 0.5969 \, \text{g}$$Rounding gives approximately $0.6 \, \text{g}$.Answer:The mass of calcium deposited is $0.6 \, \text{g}$, which matches option C.