Q38 / 180 · KCET 2024 · Chemistry

A current of 3 A is passed through a molten calcium salt for 1 hr 47 min 13 s . The mass of calcium deposited is (Molar mass of $\mathrm{Ca}=40 \mathrm{~g} \mathrm{~mol}^{-1}$ )

  • A
    6.0 g
  • B
    2.0 g
  • C
    8.0 g
  • Correct answer
    D
    4.0 g

Solution

Given, $t=1 \mathrm{hr} 47 \mathrm{~min} 13 \mathrm{sec}=6433$ seconds Molar mass of $\mathrm{Ca}=40 \mathrm{~g} / \mathrm{mol}$ Current $(i)=3 A$ Mass of calcium deposited $(w)=$ ? Using Faraday's $2^{\text {nd }}$ law $w=z i t$ ....(i) where, $z=\frac{M}{n \times 96500}=\frac{40}{2 \times 96500}$ Substitute the value of $Z$ and given values in equation (i) $W=\frac{40 \times 3 \times 6433}{2 \times 96500}=3.99 \mathrm{~g} \approx 4 \mathrm{~g}$

Hence, the correct answer is option: 'D'.

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