Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

100 mL of H2O2 solution having volume strength 11.35 V is mixed with 500 mL of given KI solution to liberate I2 gas. All the I2 gas liberated is dissolved to form 500 mL solution. 200 ml of the solution required 50 mL of  23M hypo solution. Calculate volume strength of remaining H2O2 mixture. Round off your answer to nearest integer.  [S.T.P;   Vm=22.7lit]

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya

answer is 1.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

2H2O2+2KII2+H2O+2KOH

I2+2Na2S2O3Na2S4O6+2NaI

V.S=MH2O2×11.35;M=1

200×N=50×23;N=16;M=112

100×1=500×112+x

x=58.33Mmoles

V.S=M×11.35=58.33600×11.35

Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

  • promsvg

    live classes

  • promsvg

    progress tracking

  • promsvg

    24x7 mentored guidance

  • promsvg

    study plan analysis

download the app

gplay
mentor

Download the App

gplay
whats app icon
personalised 1:1 online tutoring