wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

How much amount of CuSO45H2O is required for liberation of 2.54 g of I2 when titrated with KI?

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

The correct option is C 4.99 g
2CuSO45H2O+4KICu2I2+2K2SO4+I2254 g+10H2O

Molecular weight of 2CuSO45H2O=[2(63.5+32+64)+10(18)]g=499 g

254 g of I2 is liberated by 499g CuSO45H2O.

2.54 g of I2 will be liberated by x g CuSO45H2O.

x=499×2.54254=4.99 g

Hence, the correct option is B

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