Q.

One litre of 0.2 M acetic acid solution is passed through four grams of activated animal charcoal. The concentration of acetic acid solution coming out of charcoal was found to be 0.15 M. Value of x/m is (x= weight of adsorbate, ,m= weight of adsorbent; assume no volume changes during the process)

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!
An Intiative by Sri Chaitanya

answer is 1.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

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

Take Free Test

Detailed Solution

I) Number of moles of CH3COOH before adsorbtion0.2=n1×11 n1=0.2;II) Number of moles of CH3COOH after adsorption0.15=n2×11 n2=0.15Number of moles of CH3COOH adsorbed = (0.2 - 0.15) = 0.05Weight of CH3COOH adsorbed (x) = 0.05 x 60 =3 gGiven that the weight of charcoal (m) = 4 g∴xm=34=0.75
Watch 3-min video & get full concept clarity
Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Books for IIT JEE
Best Books for NEET
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Books for IIT JEE
Best Books for NEET
whats app icon
One litre of 0.2 M acetic acid solution is passed through four grams of activated animal charcoal. The concentration of acetic acid solution coming out of charcoal was found to be 0.15 M. Value of x/m is (x= weight of adsorbate, ,m= weight of adsorbent; assume no volume changes during the process)