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1 g charcoal is placed in 100 mL of 0.5 ...

`1 g` charcoal is placed in `100 mL` of `0.5 M CH_(3)COOH` to form an adsorbed mono-layer of acetic acid molecule and thereby the molarity of `CH_(3)COOH` reduces to `0.49`. Calculate the surface area of charcoal adsorbed by each molecule of acetic acid. Surface are of charocal `=3.01xx10^(2)m^(2)//g`.

Text Solution

Verified by Experts

The correct Answer is:
`5 xx 10^(-9) m^(2)`
Final molarity `= 5 - 49 = 01 M`
mole `= M xx v = 0.1 xx (100)/(1000)`
`= 10^(-3)`
no of molecule = moles `xx N_(A)`
`= 10^(-3) xx N_(A) = 6.02 xx 10^(20)`
1gm contain charcoal `= 3.01 xx 10^(2) m^(2)`
`6.02 xx 10^(20)` molecule of acetic acid absorbed charcoal `= 3.01 xx 10^(2)`
1 molecule of acetic acid adsorbed charcoal `= 3.01 xx 10^(2) m^(2)`
`6.01 xx 10^(20)` molecule of acetic acid absorbed charcoal `= 3.01 xx 10^(2)`
molecule of acetic acid adsorbed charcoal `= (3.01 xx 10^(2))/(6.02 xx 10^(20)) = 5 xx 10^(-19)`
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1 g of charcoal adsorbs 100 ml of 0.5 M CH_(3)COOH to form a monolyer, and thereby the molarity of CH_(3)COOH reduces to 0.49 M. Calculate the surface area of the charcoal adsorbed by each molecular of accetic acid. Surface area of charcoal =3.01xx10^(2)m^(2)g.

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Knowledge Check

  • Ig of charcoal adsorbs 100 mL 0.5 MCH_(3)COOH to form a monolayer, and thereby the molarity of CH_(3)COOH reduces to 0.49. Calculate the surface area of the charcoal covered by each molecule of acetic acid. Surface area of charcoal= 3.01 xx 10^(2) m^(2) //g

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    B
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    A
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    B
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    D
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  • 2.0 g of charcoal is placed in 100 mL of 0.5 M CH_3 COOH to form an adsorbed mono-acidic acid molecules and thereby, the molarity of CH_3 COOH reduces to 0.49 M . The surface area of charcoal is 3 xx 10^2 m^2 g^-1. The surface area of charcole adsorded by each molecule of acetic acid is ("Take" N_A=6 xx 10^23)

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    B
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    C
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