Question

# Read the passage given below and answer the question:Chemical reactions involve the interaction of atoms and molecules. A large number of atoms/ molecules (approximately $\text{6}.0\text{23 }\times \text{ 1}{{0}^{\text{23}}}$) are present in few grams of any chemical compound varying with their atomic/molecular masses. To handle such large numbers conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry and radiochemistry. The following example illustrates a typical case, involving chemical / electrochemical reaction, which requires a clear understanding of the mole concept.A $4.0$ molar aqueous solution of $NaCl$ is prepared and $500\text{ }mL$ of this solution is electrolyzed. This leads to the evolution of chlorine gas at one of the electrodes (atomic mass: $Na=23,\text{ }Hg=200;\text{ }1\text{ }faraday=96500\text{ }coulombs$ ).If the cathode is a H(g) electrode, the maximum weight (g) of amalgam formed from this solution is:A ) $200$ B ) $225$C ) $400$D ) $446$

Hint: One mole of a substance contains Avogadro’s number $6.023 \times {10^{23}}$ of molecules. To calculate the number of moles, multiply the molarity with volume (in L).

Cathode is mercury electrode. At cathode, sodium cation will gain an electron and form sodium metal. Sodium will form amalgam with mercury. Write a balanced half reaction for the formation of sodium amalgam.

$\text{N}{{\text{a}}^{+}}\text{ + 1}{{\text{e}}^{-}}\text{ + Hg }\to \text{ Na}-\text{Hg }$

$500\text{ }ml\text{ }\left( or\text{ }0.500\text{ }L \right)\text{ }of\text{ }4.0$ molar aqueous solution of sodium chloride is prepared.
To calculate the number of moles, multiply the molarity with volume (in L).
$500\text{ }ml\text{ }\left( or\text{ }0.500\text{ }L \right)\text{ }of\text{ }4.0$ molar aqueous solution of sodium chloride contains $4.0\text{ mol/L }\times \text{ 0}\text{.500 L = 2}\text{.0}$ moles of sodium chloride. This contains $2.0$ moles of sodium cations. They will form $2.0$ moles of sodium amalgam.
The atomic mass of $Na=23$ . The atomic mass of $Hg=200$ .
Calculate the mass of sodium amalgam formed.
The mass of $2.0$ moles of sodium amalgam is $2\left( 23+200 \right)=446\text{ }g$.

Hence, option D ) $446$ is the correct answer.

Note: The relationship, “$1\text{ }faraday=96500\text{ }coulombs$ ” is not used in this reaction. This relationship can be used for the problems, in which the number of electrons participating in a half reaction are given and the number or moles needs to be calculated.