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Q.

526.3 mL of 0.5 m HCI is shaken with 0.5 g of activated charcoal and filtered. The concentration of the filtrate is reduced to 0.4 m. The amount of adsorption (x/m) is

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answer is 4.

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Detailed Solution

Mass of HCI adsorbed by 10 g charcoal =526.3×10−3(0.5−0.4)×38≈2(Mw of HCI = 38 g mol-1)The amount of adsorptionxm=20.5=4
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