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

Answer:

33.67 mg $O_{2}$

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:

$m=(0.00247 \frac{mol O_{2} }{L} )*(0.426 L)*(\frac{32 g mol O_{2}}{1mol mol O_{2}} )*(\frac{1000 mg \\O_{2}}{1 g O_{2}} )$

$m=33.67 mg O_{2}$

Best regards!