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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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One gram of charcoal adsorbs 100 mL of 0.5 M CH_(3)COOH to form a mono-layer and thereby the molarity of acetic acid is reduced to 0.49 M. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid. Surface acid of charcoal =3.01xx10^(2) m^(2)//gm

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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

    A
    `0.5 xx 10^(-19) m^(2)`
    B
    `4xx 10^(-20) m^(2)`
    C
    `0.5 xx 10^(-10) m^(2)`
    D
    `5 xx10^(-19) m^(2)`

  • 2.0 g of charcoal is placed in 100 mL of 0.05 M CH_(3)COOH to form an adsorbed mono-acidic layer of acetic acid molecules and thereby the molarity of CH_(3)COOH reduces to 0.49 . The surface area of charcoal is 3xx10^(2) m^(2)g^(-1) . The surface area of charcoal is adsorbed by each molecule of acetic acid is a. 1.0xx10^(-18)m^(2) b. 1.0xx10^(-19)m^(2) c. 1.0xx10^(13)m^(2) d. 1.0xx10^(-22)m

    A
    `1.0xx10^(-18)m^(2)`
    B
    `1.0xx10^(-19)m^(2)`
    C
    `1.0xx10^(13)m^(2)`
    D
    `1.0xx10^(-22)m`

  • 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)

    A
    `1.0 xx 10^(-18) cm^2`
    B
    `1.0 xx 10^(-19) cm^2`
    C
    `1.0 xx 10^(13) cm^2`
    D
    `1.0 xx 10^(-14) cm^2`