Redox Reactions And Electrochemistry Ques 318

Question: The amount of silver deposited by passing 241.25 coulomb of current through silver nitrate solution is [MHCET 2003]

Options:

A) 2.7 g

B) 2.7 mg

C) 0.27 g

D) 0.54 g

Show Answer

Answer:

Correct Answer: C

Solution:

Given, Current = 241.25 columb 1 coulomb current will deposite $ =1.118\times {10^{-3}}gm\ Ag $ .
$ \therefore $ 241.25 current will deposite = $ 1.118\times {10^{-3}}\times 241.25 $ $ =0.27\ gm $ silver.



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक