Question

Calculate the amount of co2 dissolved at 4 atm in 1 dm3 of water at 298 k. The henry constant for co2 is 1.67 kbar

Answer 1

according to Henry's law, $P_{co_2}=K_Hx$ where x is mole fraction of CO2 present in mixture of CO2 and water .

where $P_{co_2}$ = 4atm = 4 × 1.013 × 10^5 Pa

= 4.052 × 10^5 pa [ where 1atm = 1.013 × 10^5

$K_H$ = 1.67 Kbar = 1670 bar

= 1670 × 10^5 bar

= 1.67 × 10^8 pa [ where 1bar = 10^5 pa ]

now, x = $\frac{P_{co_2}}{K_H}$

= 4.052 × 10^5/1.67 × 10^8

= 2.4 × 10^-3

= 0.0024

given, volume of water = 1dm³ = 1000cm³

we know, density of water = 1g/cm³

so, mass of water = 1000g

then mole of water = 1000/18 = 55.55 mol

now, mole fraction of CO2 = mole of CO2/(mole of CO2 + mole of Water)

or, 0.0024 = n/(n + 55.55)

or, n = 0.1336 mol

so, mole of CO2 = weight of CO2/molar weight of CO2

or, 0.1336 = weight of CO2/44g/mol

or, weight of CO2 = 0.1336 × 44 g

= 5.8784g

hence, amount of CO2 = 5.8784g

Answer 2

Answer:

From Henry law P = KH X C O 2 CO2 0.835 = 1.67 x 103 x 1.67 x 103x W C O 2 / 44 W C O 2 44 + 1000 18 WCO2/44WCO244+100018 W C O 2 WCO2 = 1.2228g = 1.222.8 x 10-3 g Or P = KHX C O 2 CO2 0.835 = 1.67 x 103 x n C O 2 n C O 2 + n H 2 O nCO2nCO2+nH2O 0.835 = 1.67 x 103 x W C O 2 / 44 1000 18 WCO2/44100018 W C O 2 CO2 = 1.222g = 1222.2 x 10-3g