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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 .

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nuhulawal20

Answer:

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)

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