Question

At 0 °C and 1.00 atm, as much as 0.70 g of O2 can dissolve in 1 L of water. At 0 °C and 4.00 atm, how many grams of O2 dissolve in 1 L of water?

Answer 1

Answer : The mass of $O_2$ at 4.00 atm pressure is, 2.8 grams.

Explanation :

First we have to calculate the Henry law constant.

As we know that,

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

where,

$C_{O_2}$ = solubility of $O_2$ = 0.70 g/L

$p_{O_2}$ = partial pressure of $O_2$ = 1.00 atm

$k_H$ = Henry's law constant

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

$0.70g/L=k_H\times (1.00atm)$

$k_H=0.70g/L.atm$

Now we have to calculate the solubility of $O_2$ at 4.00 atm pressure.

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

where,

$C_{O_2}$ = solubility of $O_2$ at 4.00 atm = ?

$p_{O_2}$ = partial pressure of $O_2$ = 4.00 atm

$k_H$ = Henry's law constant = $0.70g/L.atm$

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

$C_{O_2}=(0.70g/L.atm)\times (4.00atm)$

$C_{O_2}=2.8g/L$

Now we have to calculate the mass of $O_2$

$\text{Mass of }O_2=\text{Solubility of }O_2\times \text{Volume of solution}$

$\text{Mass of }O_2=2.8g/L\times 1L=2.8g$

Thus, the mass of $O_2$ at 4.00 atm pressure is, 2.8 grams.