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One gram of charcoal adsorbs 100 mL of 0...

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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The correct Answer is:
`5xx10^(-19)m^(2)`
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One gram of charcoal adsorbs 400 " mL of " 0.5 M acetic acid to form a mono layer and the molarity of acetic acid reduced to 0.49. Calculate the surface area of charcoal adsorbed by each molecule of acetic acid. The surface area of charcoal is 3.01xx10^(2)m^(2)g^(-1) .

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 .

Adsorption is the presence of excess concentration of any particular component at the surface of liquid a solid phase as compared to bulk.This is due to presence of residual forces at the surface of body.In the adsorption of hydrogen gas over a sample of charcoal, 1.12cm^3 of H_2(g) measured over S.T.P was found to absorb per gram of charcoal.Consider only monolayer adsorption.Density of H_2 is 0.07gm/cc. In another experiment same 1 gm charcoal adsorbs 100 ml of 0.5M CH_3COOH to form monolayer and thereby the molarity of CH_3COOH reduces to 0.49. ((3xx47.43)/(4pi))^(1//3)=2.24 (2.24)^2=5 Surface area of charcoal adsorbed by each molecule of (acetic soda) is-

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

Adsorption is the presence of excess concentration of any particular component at the surface of liquid a solid phase as compared to bulk.This is due to presence of residual forces at the surface of body.In the adsorption of hydrogen gas over a sample of charcoal, 1.12cm^3 of H_2(g) measured over S.T.P was found to absorb per gram of charcoal.Consider only monolayer adsorption.Density of H_2 is 0.07gm/cc. In another experiment same 1 gm charcoal adsorbs 100 ml of 0.5M CH_3COOH to form monolayer and thereby the molarity of CH_3COOH reduces to 0.49. ((3xx47.43)/(4pi))^(1//3)=2.24 (2.24)^2=5 Specific surface are of charcoal is (cm^2//gm) -

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)`

  • Adsorption is the presence of excess concentration of any particular component at the surface of liquid a solid phase as compared to bulk.This is due to presence of residual forces at the surface of body.In the adsorption of hydrogen gas over a sample of charcoal, 1.12cm^3 of H_2(g) measured over S.T.P was found to absorb per gram of charcoal.Consider only monolayer adsorption.Density of H_2 is 0.07gm/cc. In another experiment same 1 gm charcoal adsorbs 100 ml of 0.5M CH_3COOH to form monolayer and thereby the molarity of CH_3COOH reduces to 0.49. ((3xx47.43)/(4pi))^(1//3)=2.24 (2.24)^2=5 Surface area of charcoal adsorbed by each molecule of (acetic soda) is-

    A
    `7.8xx10^(-19) cm^2`
    B
    `7.8xx10^(-18) cm^2`
    C
    `7.8xx10^(-17) cm^2`
    D
    `7.8xx10^(-16) cm^2`

  • 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`