Henry's law constant for `CO_2` in water is `1.67 xx 10^8` 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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`K_H = 1.67 xx 10^8 Pa, P_(CO_2)` = 2.5 atm = `2.5 xx 101325 Pa` Applying Henry.s law, `P_(CO_2) = K_H xx x_(CO_2) " or " x_(CO_2) = (P_(CO_2))/(K_H)` Substituting the values, we get ` x_(CO_2) = (2.5 xx 101325 Pa)/(1.67 xx 10^8 Pa) = 1.517 xx 10^(-3)` `(n_(CO_2))/(n_(H_2O) + n_(CO_2)) = (n_(CO_2))/(n_(H_2O))= 1.517 xx 10^(-3)` For 500 mL of soda water, water present ` = 500/18 = 27.78 `moles i.e., `n_(H_2O) = 27.78` moles ` therefore (n_(CO_2))/(27.78) = 1.517 xx 10^(-3)` `n_(CO_2) = 27.78 xx 1.517 xx 10^(-3) `mole = `42.14 xx 10^(-3) xx 44 g = 1.854 g `
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U-LIKE SERIES-SOLUTION-SELF ASSESSMENT TEST (SECTION D)
