Question

At \(T(K)\), the partial pressure of dissolved oxygen in \(1\ L\) water is \(1\ bar\). The concentration of oxygen in ppm is \((K_H\) of \(O_2\) at \(T(K)\) is \(50\ kbar)\)

• A. 71.0
• B. 35.50
• C. 17.75
• D. 81.10

Hint:

Henry's law is \[ p=K_Hx \] For dilute aqueous solutions, \[ 1\ ppm \approx 1\ mg\ L^{-1} \] First calculate mole fraction using Henry's law, then convert dissolved gas into mass per litre.

Answer 1

The Correct Option is B

Step 1: Use Henry's law.
According to Henry's law, \[ p=K_Hx \] where \(p\) is the partial pressure of gas, \(K_H\) is Henry's law constant and \(x\) is the mole fraction of the gas in solution.
Given, \[ p=1\ bar \] \[ K_H=50\ kbar=50\times 10^3\ bar \]

Step 2: Calculate mole fraction of oxygen.
\[ x=\frac{p}{K_H} \] \[ x=\frac{1}{50\times10^3} \] \[ x=2\times10^{-5} \]

Step 3: Calculate moles of water in \(1\ L\).
For \(1\ L\) water, \[ \text{Mass of water}=1000\ g \] Molar mass of water is \[ 18\ g\ mol^{-1} \] Hence, \[ n_{H_2O}=\frac{1000}{18} \] \[ n_{H_2O}=55.55 \]

Step 4: Calculate moles and mass of oxygen.
For dilute solutions, \[ x_{O_2} = \frac{n_{O_2}}{n_{H_2O}} \] Therefore, \[ n_{O_2}=x_{O_2}\times n_{H_2O} \] \[ n_{O_2}=2\times10^{-5}\times55.55 \] \[ n_{O_2}=1.111\times10^{-3} \] Mass of oxygen dissolved is \[ m=n\times M \] \[ m=1.111\times10^{-3}\times32 \] \[ m=0.03555\ g \] \[ m=35.55\ mg \]

Step 5: Convert into ppm.
For aqueous solutions, \[ 1\ ppm \approx 1\ mg\ L^{-1} \] Since the solution volume is \(1\ L\), \[ \text{Concentration}=35.55\ ppm \] Approximately, \[ 35.50\ ppm \]

Step 6: Final conclusion.
Hence, \[ \boxed{35.50} \] Therefore, option (2) is correct.