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

# 0.1 g KIO3 and excess KI when treated with HCl, the iodine is liberated. The liberated iodine required 45 mL sodium thiosulphate for titration. The molarity of sodium thiosulphate will be equivalent to:

A
0.252 M
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
0.126 M
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
0.0313 M
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
0.0623 M
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

## The correct option is D 0.0623 MGiven,0.1gKIO3, molar mass of KIO3=214∴ Molar of KIO3=GivenmassMolarmass=0.1214=0.00047molesNow, KIO3+5KI+3H2SO4⇌3K2SO4+3H2O+3I2↑I2 is liberated from KIO3,⇒0.00047÷2=0.000235Now, moles of KI reacting =0.00047×5=0.000235∴ moles of I2 Produced from KI=0.000235÷2=0.0001175∴ Total moles of I2=0.000235+0.0001175=0.000141 equivalent of I2Equivalent of Thiosulphate =2×0.000141=0.000282∴ Morality =0.000282×100045=0.00626M

Suggest Corrections
1
Join BYJU'S Learning Program
Related Videos
Acids and Bases
CHEMISTRY
Watch in App
Explore more
Join BYJU'S Learning Program