0

Guest

Courses New

KHC 2 O 4 solution is 4N when treated with NaOH. What is the normality of this solution when oxidised by KMnO 4 ? Please explain in detail....

KHC2O4 solution is 4N when treated with NaOH. What is the normality of this solution when oxidised by KMnO4 ?
Please explain in detail....
 
 

Grade:12th pass

2 Answers

Vikas TU
14149 Points
5 years ago
Equivalents of KMnO4 =  Equivalents of KHC2O4
KMnO4 oxidezed into Mn2+ and KHC2O4 into CO2.
Valence factor of KMnO4 therefore,= 7 – 5 = +5
Valence factor of KHC2O4 = 2
4*2 = N*5
N = 8/5 => 1.6.
Yasir Rather
31 Points
5 years ago
Please Vikas TU sir explain..............
eq of KMnO4 = Eq of KHC2O4
NV = NV
M *n-factor *V = M *n-factor *V
no. of moles* n-factor= no. of moles* n-factor

Think You Can Provide A Better Answer ?

Other Related Questions on Physical Chemistry

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More