Question

During the preparation of $H_2{S}_2{O}_8$ (per disulphuric acid) $O_2$ gas also releases at anode as by product. When $9.72L$ of $H_2$ releases at cathode and $2.35L{O}_2$ at anode at STP, the weight of $H_2{S}_2{O}_8$ produced in gram is:

• A. $87.12$
• B. $43.56$
• C. $83.42$
• D. $51.74$

Answer 1

Answer: (B)

The explanation for the correct option:

Option B $43.56$

• ​By faraday's Law, the equivalence of hydrogen is equal to the sum of equivalence of oxygen and per disulphuric acid.

$eq.of{H}_2=eq.of{O}_2+eq.of{H}_2{S}_2{O}_8$

$\frac{9.85}{\left(\frac{22.4}{2}\right)}=\frac{2.35}{\left(\frac{22.4}{4}\right)}+\frac{W}{\left(\frac{194}{2}\right)}$

$\frac{9.85}{11.2}=\frac{2.35}{5.6}+\frac{W}{97}$

$\frac{W}{97}=\frac{9.85}{11.2}-\frac{2.35}{5.6}$

$=\frac{28.84}{62.72}$

$\therefore W=\frac{28.84}{62.72}\times 97=43.56gm$

Hence, Option B 43.56gm is correct