Q4 / 90 · KVPY 2016 (SB/SX) · Chemistry

For electroplating, $1.5$ amp current is passed for 250 s through $250 \mathrm{mI}$. of $0.15 \mathrm{M}$ solution of $\mathrm{MSO}_{4}$. Only $85 \%$ of the current was utilized for electrolysis. The molarity of $\mathrm{MSO}_{4}$ solution after electrolysis is closest to [Assume that the volume of the solution remained constant]

  • Correct answer
    A
    $0.14$
  • B
    $0.014$
  • C
    $0.07$
  • D
    $0.035$

Solution

No. of mole $=\frac{\mathrm{i} \times \mathrm{t}}{96500 \times \mathrm{v} . \mathrm{f}}, \quad \mathrm{n} \times \mathrm{v} . \mathrm{f}=\frac{\mathrm{i} \times \mathrm{t}}{96500} \times \mathrm{n}=\frac{1.5 \times 250}{96500 \times 2} \times 0.85$ $\mathrm{n}=.00165$ (deposited) $\mathrm{n}_{\mathrm{i}}=\frac{250 \times .15}{1000}=.0375$ (initial mole) $\mathrm{n}_{\text {left }}=.03585$ (mole left) $\mathrm{M}_{\text {left }}=\frac{.03585}{.25}=0.143$

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

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