Question

The Henry’s law constant for N2 in water is 1*10^5 atm. Assuming that air contains 80 %N2,the number of moles of N2 dissolved in 10^5 moles of water at 5 atm pressure is

Answer 1

Answer:pressure of N2=5×0.8=4

Applying henry's law

P=KH.partial pressure of nitrogen

4=1×10^5×partial pressure of N2

So,

Partial pressure of N2(xn2)=4×10^-5

Xn2=no.of moles of N2/no.of moles of N2+no.of moles of H2O

4×10^-5=nN2/nN2+10^5

So, by solving

nN2=4

Explanation:therefore no.of moles of nitrogen dissolved in 10^5 moles of water at 5 atm pressure =4