Question

What mass of oxygen gas, in mg, is dissolved in 4.26×102 mL of water when the partial pressure of oxygen above the water is 1.9 atm at 25 °C? The Henry's constant for oxygen gas in water at 25 °C is 0.0013 M/atm.

Answer 1

33.67 mg
`O_(2)`

Step-by-step explanation:

Hello,

Since the Henry's constant allows us to know the concentration of the dissolved
`O_(2)`, we apply the formula:

`C=P*H\\C=1.9 atm *   0.0013 M/atm\\C= 0.00247 M`

Now, by knowing the concentration and that the volume is 0.426 L of water (proper units), we apply the following factors to know the milligrams of oxygen gas:

`m=(0.00247 (mol O_(2) )/(L) )*(0.426 L)*((32 g mol O_(2))/(1mol mol O_(2)) )*((1000 mg \\O_(2))/(1 g O_(2)) )\\`

`m=33.67 mg O_(2)`

Best regards!