CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

A drop of solution (volume 0.05 mL) contains 3×106 mole of H+. If the rate constant of the disappearance of H+ is 107mollitre1sec1, how long would it take for H+ in the drop to disappear?


Open in App
Solution

Step 1: Given data

Volume of solution= 0.05mL

Mole of H+= 3×106 mole

The rate constant of the disappearance of H+= 107mollitre1sec1

Step 2: Calculating the concentration of drop

We know that the concentration of the dop is

Concentrationofdrop=MolesofthesolutionVolume(inmL)×1000Concentrationofdrop=3×10-60.05×1000Concentrationofdrop=0.06molL-1

Step 3: Calculating the time

Let us assume the time taken for theH+ in the drop to disappear be T

Now we know that the rate of disappearance is

Rateofdisappearance=ChangeinconcentrationTime(T)Time(T)=ChangeinconcentrationRateofdisappearanceTime(T)=0.06107Time(T)=6×10-9sec

Therefore, the time taken for the H+ in the drop to disappear is 6×10-9sec.


flag
Suggest Corrections
thumbs-up
2
BNAT
mid-banner-image