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प्रश्न
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.
उत्तर
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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