Electrolysis of hot aqueous solution of NaCl gives NaClO4 , i.e., sodium perchlorate, NaCl+4H2O arrow NaClO4+4H2 uparrow How many Faraday are required to obtain 1000 g of sodium perchlorate? Report your answer by rounding it up to nearest whole number.

Question

Electrolysis of hot aqueous solution of NaCl gives NaClO4 , i.e., sodium perchlorate, NaCl+4H2O arrow NaClO4+4H2 uparrow How many Faraday are required to obtain 1000 g of sodium perchlorate? Report your answer by rounding it up to nearest whole number. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Equivalent mass of NaClO4=( textMolecular mass/ textChange in oxidation state of chloride) =(122.5/+ 7 - (. - 1 .))=15.13 Number of Faraday = Number of equivalent of NaClO4 formed =(1000/15.13)=66

Q. Electrolysis of hot aqueous solution of $N a Cl$ gives $N a Cl O_{4}$ , i.e., sodium perchlorate,
$N a Cl + 4 H_{2} O \to N a Cl O_{4} + 4 H_{2} ↑$
How many Faraday are required to obtain $1000 g$ of sodium perchlorate? Report your answer by rounding it up to nearest whole number.

Equivalent mass of $N a Cl O_{4} = \frac{Molecular mass}{Change in oxidation state of chloride}$
$= \frac{122.5}{+ 7 - ( - 1 )} = 15.13$
Number of Faraday = Number of equivalent of $N a Cl O_{4}$ formed
$= \frac{1000}{15.13} = 66$

Q. Electrolysis of hot aqueous solution of NaCl gives NaClO4​ , i.e., sodium perchlorate, NaCl+4H2​O→NaClO4​+4H2​↑ How many Faraday are required to obtain 1000g of sodium perchlorate? Report your answer by rounding it up to nearest whole number.

Equivalent mass of NaClO4​=Change in oxidation state of chlorideMolecular mass​ =+7−(−1)122.5​=15.13 Number of Faraday = Number of equivalent of NaClO4​ formed =15.131000​=66

Q. Electrolysis of hot aqueous solution of $N a Cl$ gives $N a Cl O_{4}$ , i.e., sodium perchlorate,
$N a Cl + 4 H_{2} O \to N a Cl O_{4} + 4 H_{2} ↑$
How many Faraday are required to obtain $1000 g$ of sodium perchlorate? Report your answer by rounding it up to nearest whole number.

Equivalent mass of $N a Cl O_{4} = \frac{Molecular mass}{Change in oxidation state of chloride}$
$= \frac{122.5}{+ 7 - ( - 1 )} = 15.13$
Number of Faraday = Number of equivalent of $N a Cl O_{4}$ formed
$= \frac{1000}{15.13} = 66$