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526.3mL of 0.5mHCl is shaken with 0.5g o...

`526.3mL` of `0.5mHCl` is shaken with `0.5g` of activated charcoal and filtered. The concentration of the filtrate is reduced to `04m` . The amount of adsorption `(x//m)` is

A

3

B

6

C

8

D

4

Text Solution

Verified by Experts

The correct Answer is:
D
Mass of `HCl` acid adsorbed by `10g` charcoal
`=526.3xx10^(-3)(0.5-0.4)xx38=2`
(Mw of `HCl=38gmol^(-1))`
The amount of adsorption
`(x)/(m)=(2)/(0.5)=4`
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