# At what partial pressure, oxygen will have solubility of 0.05 g L-1 in water at 293 K? Henry’s constant (KH) for O2 in water at 293 K is 34.36 k bar. Assume the density of the solution to be same as that of the solvent.

**At what partial pressure, oxygen will have solubility of 0.05 g L ^{-1} in water at 293 K? Henry’s constant (K_{H}) for O_{2} in water at 293 K is 34.36 k bar. Assume the density of the solution to be same as that of the solvent.**

**Answer:**

**Given Values**

A = H_{2}O , B = O_{2}

K_{H} = 34.36 k bar = 34.36 × 10^{3} bar

Solubility = 0.05 gm/L

∴ W_{B} = 0.05 gm , V_{sol} = 1 L or 1000 ml

M_{B} = 2(16) = 32

D_{sol} = D_{water (A)} = 1 gm/ml

∴ W_{sol} = 1000 gm

W_{A} = W_{sol} − W_{B} = 1000 − 0.005 ≈ 1000 gm

M_{A} = 2+16 = 18

**Calculations**

**Calculate mole of Oxygen (n _{B})**

\( n_{B} = \frac{W_{B}}{M_{B}} = \frac{0.05}{32} = 0.0016 \) moles

**Calculate mole of Solvent or Water (n _{A})**

\( n_{A} = \frac{W_{A}}{M_{A}} = \frac{1000}{18} = 55.5 \) moles

**Calculate Partial pressure of Oxygen**

\( P = K_{H}\frac{n_{B}}{n_{A}+n_{B}} \)

\( P = 34.36 \times 10^{3}~ \frac{0.0016}{55.5+0.0016} \)

**P = 0962 bar**

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