3 g of activated charcoal was added to 50 mL 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 adsorbed (per gram of charcoal) is

Question

3 g of activated charcoal was added to 50 mL 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 adsorbed (per gram of charcoal) is (A) 18 mg (B) 36 mg (C) 42 mg (D) 54 mg

Tardigrade Admin · Accepted Answer

CH 3 COOH (0.06 M ) 50 ml m. moles =50 × 0.06=3 m. moles left =50 × 0.042=2.1 m. moles absorbed =0.9 mass absorbed =(0.9 × 10-3 × 60/3) × 103 =(54/3)=18 mg

Q. $3 g$ 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 adsorbed (per gram of charcoal) is

18 mg

36 mg

42 mg

54 mg

$C H_{3} COO H (0.06 M)$

$50 m l$

$m$ . moles $= 50 \times 0.06 = 3$

$m$ . moles left $= 50 \times 0.042 = 2.1$

$m$ . moles absorbed $= 0.9$

mass absorbed $= \frac{0.9 \times 1 0 ^{- 3} \times 60}{3} \times 1 0^{3}$

$= \frac{54}{3} = 18 m g$

Q. 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 adsorbed (per gram of charcoal) is