Question

Calculate the solubility of Co2 in water at 298 Kelvin under 760 mm Hg

(Kh for CO2 =
$1.25 \times 10 ^{6}$
​

Answer 1

the solubility of CO₂ in water at 298 Kelvin under 760 mm Hg  is 0.0337 mol/litre

- By using Henry's law

P = $K_{H}$ × x

x = P / $K_{H}$

  = 760 / (1.25 × 10⁶)

  = 608 × 10⁻⁶

  = 6.08 × 10⁻⁴

As mole fraction is related to solubility, hence solubility in terms of mole fraction is 6.08 × 10⁻⁴

* In terms of mol / litre, solubility can be calculated as follows :

- mole fraction,

x = no.of moles of solute/ total no. of moles in solution

x = no.of moles of solute/ ( no.of moles of solute + no. of moles of water in 1 litre)

                         x = n / (n+55.55)

As, moles of CO₂ <<<< moles of wate

                         x ≈ n / 55.55

-    6.08 × 10⁻⁴ = n / 55.55

                n = 6.08 × 10⁻⁴ × 55.55

                n = 6.08 × 10⁻⁴ × 55.55

                n = 337.74 × 10⁻⁴

                n = 0.0337

- so solubility = 0.0337 mol/litre

Answer 2

The solubility of CO2 in water is 33.74 x 10^-3 mol/L.

Explanation:

We are given that:

• Temperature "T" = 298 K  
• kH = 1.25 x 10^6  
• Pressure "p" = 760 mmHg
• Henry's Law states that  
• Pressure = kH x (x)      

Where "x" is the mole fraction of the gas in the solution.

760 = 1.25  x 10^6 x (x)  

x = 760 / 1.25 x 10^6 ,      

x = 608 x 10^6  

Now, we know mole fraction of gas                                                                         x = moles of gas/moles of gas + moles of water

608 x 10^6 = moles of gas / moles of gas + 55.5      

Moles of gas is less than moles of water  

Moles of gas = 608 x 10^-6 x 55.5 mol = 33744 x 10^10^-6  = 33.74 x 10^-3  mol/L

Thus the solubility of CO2 in water is 33.74 x 10^-3 mol/L.