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Henry’s law constant for CO2 in water is 1.67 × 108 Pa at 298 K. Calculate the quantity of CO2 in 500 mL of soda water when packed under 2.5 atm CO2 pressure at 298 K. - Chemistry

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Question

Henry’s law constant for CO2 in water is 1.67 × 108 Pa at 298 K. Calculate the quantity of CO2 in 500 mL of soda water when packed under 2.5 atm CO2 pressure at 298 K.

Numerical

Solution

Given, KH = 1.67 × 108 Pa

`"P"_("CO"_2)` = 2.5 atm = 2.5 × 101325 Pa

According to Henry’s law,

`"P"_("CO"_2) = "K"_"H" xx "x"_("CO"_2)`

∴ `"x"_("CO"_2) = "P"_("CO"_2)/"K"_"H"`

= `(2.5 xx 101325  "Pa")/(1.67 xx 10^8  "Pa")`

= 1.517 × 10−3

We can write, `"x"_("CO"_2) = ("n"_("CO"_2))/("n"_("CO"_2) + "n"_("H"_2"O")) = ("n"_("CO"_2))/("n"_("H"_2"O")) = 1.517 × 10^(−3)  ...["Since, n"_("CO"_2)  "is negligible as compared to n"_("H"_2"O")]`

For 500 mL of soda water, the volume of water = 500 mL; the mass of water = 500 g,

We can write:

= `500/18  "mol of water"`

= 27.78 mol of water

i.e., `"n"_("H"_2"O")` = 27.78

∴ `("n"_("CO"_2))/"n"_("H"_2"O") = "x"_("CO"_2)`

`("n"_("CO"_2))/(27.78) = 1.517 xx 10^(−3)`

or `"n"_("CO"_2)` = 42.14 × 10−3 mole

Mass of CO2 = 42.14 × 10−3 × 44 g = 1.854 g

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Chapter 2: Solutions - Intext Questions [Page 41]

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NCERT Chemistry [English] Class 12
Chapter 2 Solutions
Intext Questions | Q 7 | Page 41

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