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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 `0.4M` . The amount of adsorption `(x//m)` is

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The correct Answer is:
4

Mass of HCl acid adsorbed by 1- g charcoal `=526.3 xx 10^(-3) (0.5 -0.4) xx 38=2` (Mw of HCl `=38 g mol^(-1)`)
The amount of adsorption.
`x/m = 2/(0.5) =4`
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