The partial pressure of ethane over a solution containing 6.56 x 10-3 g of ethane is 1 bar. If the solution contains 5.00 x 10-2 g of ethane, then what shall be the partial pressure of the gas?
According to Henry’s law
.m = k x p
Substituting the given values in the above equation.
We get
6.56 x10-3 = k x 1
Or
k = 6.56 x10-3
Now when m = 5 x 10-2,
Then again substituting the given values in Henry’s law equation, we get
5 x 10-2 = 6.56 x10-3 x p
Or
p = 7.62 bar
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(i) Propanal and Propanone
(ii) Acetophenone and Benzophenone
(iii) Phenol and Benzoic acid
(iv) Benzoic acid and Ethyl benzoate
(v) Pentan-2-one and Pentan-3-one
(vi) Benzaldehyde and Acetophenone
(vii) Ethanal and Propanal
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How the following conversions can be carried out?
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(ii) Ethanol to but-1-yne
(iii) 1-Bromopropane to 2-bromopropane
(iv) Toluene to benzyl alcohol
(v) Benzene to 4-bromonitrobenzene
(vi) Benzyl alcohol to 2-phenylethanoic acid
(vii) Ethanol to propanenitrile
(viii) Aniline to chlorobenzene
(ix) 2-Chlorobutane to 3, 4-dimethylhexane
(x) 2-Methyl-1-propene to 2-chloro-2-methylpropane
(xi) Ethyl chloride to propanoic acid
(xii) But-1-ene to n-butyliodide
(xiii) 2-Chloropropane to 1-propanol
(xiv) Isopropyl alcohol to iodoform
(xv) Chlorobenzene to p-nitrophenol
(xvi) 2-Bromopropane to 1-bromopropane
(xvii) Chloroethane to butane
(xviii) Benzene to diphenyl
(xix) tert-Butyl bromide to isobutyl bromide
(xx) Aniline to phenylisocyanide
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A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate:
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2) Vapour pressure of water at 298 K.
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Explain the fact that in aryl alkyl ethers
(i) The alkoxy group activates the benzene ring towards electrophilic substitution and
(ii) It directs the incoming substituents to ortho and para positions in benzene ring.
Henry's law constant for CO2 in water is 1.67 x 108Pa at 298 K. Calculate the quantity of CO2in 500 mL of soda water when packed under 2.5 atm CO2 pressure at 298 K.
Hey frnds this ans is correct but this is not the correct way to solve. To solve this make two equation taking no. Of moles of other component in solution as n and Henry's constant as k and partial pressure of ethane in second solution as x .While solving both equations Henry's constant and no. Of moles of other component get cancelled and u get x which is required ans.
Will you explain it in a more elaborate way?
p pressure is directly proportional to mole fraction of gas then how it becomes mass??? can anybody explain???
Partial pressure is directly proportional to the mass of the gas
P=Kh*x....where x is mole fraction then how we can replace it by mass?.... ðð
Isn't it kh*m=p. They have given it in the opposite way
I could not undrstnd that Henry\'s law states that partial pressure in vapour phase is directly proportinal to mole fraction ,then how Can it be apply directly to the given mass by replacing the mole fraction..? That is done here
Can we solve it with pie = CRT?