Home
Class 12
CHEMISTRY
One gram of charcoal adsorbs 100ml of 0....

One gram of charcoal adsorbs `100ml` of `0.5M` acetic acid to form a monolayer, and the molarity of acetic acid reduces 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)` .

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)`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • CONCENTRATION TERMS

    ALLEN|Exercise Basic Exercise|14 Videos

  • CONCENTRATION TERMS

    ALLEN|Exercise Exercise S - I|30 Videos

  • CONCENTRATION TERMS

    ALLEN|Exercise Exercise Jee-Advance|3 Videos

  • ACIDIC STRENGTH & BASIC STRENGTH

    ALLEN|Exercise Exercise V|16 Videos

Similar Questions

Explore conceptually related problems

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 .

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 Surface area of charcoal adsorbed by each molecule of (acetic soda) is-

5 mL of 0.3 M acetic acid is shaken with 5 g of activated charcoal. The concentration of acetic acid is reduced to 1/3 of the original molarity. The weight of acetic acid adsorbed per gram of charcoal is:

100 mL of 0.3 M acetic acid is shaken with 0.8 g wood charcoal. The final concentration of acetic acid in the solution after adsorption is 0.125 M. The mass of acetic acid adsorbed per gram of charcoal is :

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 adsorbed (per gram of charcoal) is:

Volume of N_(2) at 1 atm, 273 K required to form a monolayer on the surface of iron catalyst is 8.15ml//gm of the adsorbent. What will be the surface area of the adsorbent per gram if each nitrogen molecule occupies 16 xx 10^(-22)m^(2) ? [Take : N_(A)=6xx10^(23) ]

The depression of freezing point of a solution of acetic acid in benzene is – 0.2°C. If the molality of acetic acid is 0.1 m, then find the ratio of the normal mass to the abnormal mass. (Assume Kf of acetic acid = 4.0°C m-1)

The density of a 1M solution of acetic acid in water is 2g/mL. The molality of the solution is

Home
Profile