$3 g$ of activated charcoal was added to $50 m L$ of acetic acid solution ( $0.06 N$ ) in a flask. After an hour it was filtered and the strength of the filtrate was found to be $0.042 N$ . The amount of acetic acid absorbed (per gram of charcoal) is ?

18 m g

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36 m g

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42 m g

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54 m g

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Solution

The correct option is A $18 m g$
Given:

The initial mass of charcoal $= 3 g$

Vol. of acetic acid = $50 m L$ Normality $= 0.06 N$

Strength of filtrate after reaction $= 0.042 N$

We know,

Molarity of acetic acid solⁿ $= \frac{W t / M}{V o l o f s o l^{n}}$

$\Rightarrow W t = 0.06 \times 60 \times 0.05 = 0.180 g$

Also, normality after reaction, $0.042 = \frac{W}{60 \times 0.05}$

$W = 0.126 g$

Change in mass $= 0.180 - 0.126 = 0.054 g = 54 m g$

Amount of charcoal available $= 3 g$

Amount of acetic acid absorbed $= \frac{54 m g}{3.0 g} \times 1.0 g = 18 g$
Correct option is A.

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3g of activated charcoal was added to 50mL of acetic acid solution (0.06N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042N. The amount of acetic acid absorbed (per gram of charcoal) is ?

$3 g$ of activated charcoal was added to $50 m L$ of acetic acid solution ( $0.06 N$ ) in a flask. After an hour it was filtered and the strength of the filtrate was found to be $0.042 N$ . The amount of acetic acid absorbed (per gram of charcoal) is ?