CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The percentage weight of Zn in white vitriol [ZnSO47H2O] is approximately equal to?

A
33.65%
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
32.56%
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
23.65%
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
22.65%
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D 22.65%
Molecular weight of white vitriol ZnSO4.7H2O=65+32+(4×16)+7(2×1+16)=287 g
Weight of Zn in white vitriol = 65 g
Percentage weight of Zn = WeightofZnMolecularweightofwhitevitriol×100=65287×100=22.65 %

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