A Drop Of Solution (Volume 0.05 Ml) Contains 3×10 −6 Mole Of H+ . If The Rate Constant Of Disappearance Of H + Is 10 7 Mol Litre −1Sec −1 , How Long Would It Take For H + In The Drop To Disappear?

The concentration of drop is given by

\( \frac{Mole}{Volume in ML} * 1000 \) \( \Rightarrow \frac{3 * 10^{-6}}{0.05L} * 1000 \) \( \Rightarrow 0.06 mol litre^{-1} \)

The rate of disappearance is given by

\( \frac{Change in concentration} {Time} \) \( \Rightarrow 1 * 10^{7} = \frac{0.06} {Time} \)

Time = \( 6 * 10^{-9}sec \)

Therefore, the total time required for \( H^{+} to disappear is 6 * 10^{-9}sec \)

Explore more such questions and answers at BYJU’S.

Was this answer helpful?

 
   

0 (0)

(0)
(0)

Choose An Option That Best Describes Your Problem

Thank you. Your Feedback will Help us Serve you better.

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *

*

*

BOOK

Free Class

Ask
Question