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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).
Complete answer:
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.
Complete answer:
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.
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