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Q.

One litre of 0.4 M acetic acid solution is passed through twelve 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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a

0.75

b

0.5

c

1.5

d

0.15

answer is A.

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Detailed Solution

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

$0.4 = n_{1} \times \frac{1}{1} n_{1} = 0.4;$

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 = (0.4 - 0.25) = 0.15

Weight of ${CH}_{3} COOH$ adsorbed (x) = 0.15 x 60 =9 g

Weight of charcoal (m) = 12 g

$∴ \frac{x}{m} = \frac{9}{12} = 0.75$

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One litre of 0.4 M acetic acid solution is passed through twelve 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)