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Henry' s law constant for CO(2) in water...

Henry' s law constant for `CO_(2)` in water is 1.67 K bar at `25^(@)C.` The quantity of `CO_(2)` in 1000 mL of soda water when packed under 5 bar `CO_(2)` pressure at `23^(@)C` is

A

0.084

B

0.167 mol

C

0.252 mol

D

0.336 mol

Text Solution

Verified by Experts

The correct Answer is:
B
We know that
`p=K_(H)X_(i)rarr p- K_(H)xx(n_(1))/(n_(1+n_(2)))`
Given is p = pressure (5 bar)
`K_(H)` = Henery. s constant (1.67 K bar.)
`X_(1)` = mole fraction of solute `CO_(2)`
`X_(1)= (n_(1))/(n_(1)+n_(2))rArr 5 = 1.67 xx 10 ^(3)xx (H_(1))/(55.5)`
`H_(1)= (555xx5)/(1.67xx10^(3))=0.167` mol
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Henry's law constant for CO_(2) in water is 1.67xx10^(8) Pa at 298 K. Calculate the quantity of CO_(2) in 500 mL of soda water when packed under 2.5 atm CO_(2) pressure at 298 K.

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Knowledge Check

  • Henry's law constant for CO_2 in water is 1.67 kbar at 25^@ C . The quantity of CO_2 in 1000 mL of soda water when packed under 5 bar CO_2 pressure at 25^@ C is

    A
    0.084 mol
    B
    0.167 mol
    C
    0.252 mol
    D
    0.336 mol

  • The quantity of CO_(2) in 500 mL of soda water when packed under 3.34 bar CO_(2) pressure at 298 K in g is

    A
    2.442
    B
    1.221
    C
    4.884
    D
    3.663

  • The quantity of CO_2 in 500ml of soda water when packed under 2.5 atm CO_2 pressure at 298 k is....... gm (Henry's law constant 1.67 xx 10^8 pa at 298K)

    A
    0.64
    B
    1.86
    C
    6.4
    D
    18.6

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