How many atoms of calcium will be deposited at the cathode when a current of $25 \ mA$ is passed through a $CaCl_2$ solution for $60 \ s$?

  • A
    $4.68 \times 10^{18}$
  • B
    $4.68 \times 10^{15}$
  • C
    $4.68 \times 10^{12}$
  • D
    $4.68 \times 10^{9}$

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Match List $I$ with List $II$.
List $I$ (Conversion)List $II$ (Number of Faraday required)
$A$. $1 \text{ mol of } H_2O \text{ to } O_2$$I$. $3F$
$B$. $1 \text{ mol of } MnO_4^- \text{ to } Mn^{2+}$$II$. $2F$
$C$. $1.5 \text{ mol of } Ca \text{ from molten } CaCl_2$$III$. $1F$
$D$. $1 \text{ mol of } FeO \text{ to } Fe_2O_3$$IV$. $5F$
Choose the correct answer from the options given below:

When an electric current is passed through acidulated water,$112 \ mL$ of hydrogen gas at $N.T.P.$ collects at the cathode in $965 \ seconds$. The current passed,in amperes,is:

Salts of $A$ (atomic weight $8$),$B$ (atomic weight $18$) and $C$ (atomic weight $50$) were electrolysed under identical conditions using the same quantity of electricity. It was found that $2.4 \ g$ of $A$ was deposited,the weight of $B$ and $C$ deposited are $1.8 \ g$ and $7.5 \ g$ respectively. The valences of $A$,$B$ and $C$ are,respectively,

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