CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

0.5 g of an oxalate was dissolved in water and the solution made to 100 mL. On titration 10 mL of this solution required 15 mL of N20KMnO4. Calculate the percentage of oxalate in the sample.

Open in App
Solution

15 mLN20KMnO4=10× Normality of oxalate solution.
Normality of oxalate solution =1520×120=340
Strength of oxalate solution =Normality× Eq. mass of oxalate
=340×44=3.3 g/L [Eq. mass of C2O24=882=44]
Amount of oxalate in 100 mL solution =3.31000×100=0.33 g
% of oxalate =0.330.5×100=66.0

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Examples of Changes
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon