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One litre of 0.3 M acetic acid solution is passed through six grams of activated animal charcoal. The concentration of acetic acid solution coming out of charcoal was found to be 0.25 M. Value of x/m is (x= weight of adsorbate, ,m= weight of adsorbent; assume no volume changes during the process)

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0.75

0.15

1.5

0.5

answer is B.

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Detailed Solution

I) Number of moles of ${CH}_{3} COOH$ before adsorbtion

$0.3 = n_{1} \times \frac{1}{1} n_{1} = 0.3;$

II) Number of moles of ${CH}_{3} COOH$ after adsorption

$0.25 = n_{2} \times \frac{1}{1} n_{2} = 0.25;$

Number of moles of ${CH}_{3} COOH$ adsorbed = (0.3 - 0.25) = 0.05

Weight of ${CH}_{3} COOH$ adsorbed (x) = 0.05 x 60 =3 g

Given that the weight of charcoal (m) = 6 g

$∴ \frac{x}{m} = \frac{3}{6} = 0.5$

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