Question

At a certain temperature, the solubility of N₂ gas in water at 2.38 atm is 56.0 mg of N₂ gas/100 g water. Calculate the solubility of N₂ gas in water, at the same temperature, if the partial pressure of N₂ gas over the solution is increased from 2.38 atm to 5.00 atm.

Answer 1

Answer: The solubility of nitrogen gas at 5.00 atm is
`117.6mg/100g`

Step-by-step explanation:

To calculate the molar solubility, we use the equation given by Henry's law, which is:

`C_(A)=K_H* p_(A)`

Or,

`(C_(1))/(C_(2))=(p_(1))/(p_2)`

where,

`C_1\text{ and }p_1` are the initial concentration and partial pressure of nitrogen gas

`C_2\text{ and }p_2` are the final concentration and partial pressure of nitrogen gas

We are given:

`C_1=56.0mg/100g\\p_1=2.38atm\\C_2=?\\p_2=5.00atm`

Putting values in above equation, we get:

`(56.0mg/100g)/(C_2)=(2.38atm)/(5.00atm)\\\\C_2=(56.0mg/100g* 5.00atm)/(2.38atm)=117.6mg/100g`

Hence, the solubility of nitrogen gas at 5.00 atm is
`117.6mg/100g`