Question

The Henry's law constant (kH) for O2 in water at 20°C is 1.28e-3 mol/l atm. How many grams of O2 will dissolve in 3.5 L of H2O that is in contact with pure O2 at 1.65 atm?

Answer 1

Answer : The mass of $O_2$ dissolved will be, 0.2365 grams

Explanation :

First we have to calculate the concentration of $O_2$.

As we know that,

$C_{O_2}=k_H\times p_{O_2}$

where,

$C_{O_2}$ = concentration of $O_2$ = ?

$p_{O_2}$ = partial pressure of $O_2$ = 1.65 atm

$k_H$ = Henry's law constant = $1.28\times 10^{-3}mole/L.atm$

Now put all the given values in the above formula, we get:

$C_{O_2}=(1.28\times 10^{-3}mole/L.atm)\times (1.65atm)$

$C_{O_2}=2.112\times 10^{-3}mole/L$

The concentration of $O_2$ = $2.112\times 10^{-3}mole/L$

Now we have to calculate the moles of $O_2$

$\text{Moles of }O_2=\text{Concentration of }O_2\times \text{volume of solution}$

$\text{Moles of }O_2=(2.112\times 10^{-3}mole/L)\times (3.5L)=7.392\times 10^{-3}mole$

Now we have to calculate the mass of $O_2$

$\text{Mass of }O_2=\text{Moles of }O_2\times \text{Molar mass of }O_2$

$\text{Mass of }O_2=(7.392\times 10^{-3}mole)\times (32g/mole)=0.2365g$

Therefore, the mass of $O_2$ dissolved will be, 0.2365 grams