At a certain temperature, the solubility of N2 N 2 gas in water at 2.38 atm a t m is 56.0 mg of N2 gas/100 g water m g o f N 2 g a s / 100 g w a t e r . Calculate the solubility of N2 N 2 gas in water, at the same temperature, if the partial pressure of N2 N 2 gas over the solution is increased from 2.38 atm a t m to 5.00 atm a t m .

solubility is 117.64 (mg/100 mL)

Explanation:

Given the data in the question;

According to Henry's Law; solubility is directly proportional to the partial pressure;

at 2.38 atm solubility of N2 is 56.0 mg of N2 gas/100 g water

if the partial pressure of N2 gas increases from 2.38atm to 5.00atm,

the solubility of N2 gas in water = ?

so

solubility at 2.38 atm = (56 × 5) / 2.38 (mg/100 mL)

= 280 / 2.38 (mg/100 mL)

= 117.64 (mg/100 mL)

Therefore, solubility is 117.64 (mg/100 mL)

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.