3 a Given an atmospheric pressure of 689 mmHg what is the pa

3. (a) Given an atmospheric pressure of 689 mmHg what is the partial pressure of nitrogen 0.7809) of the air molecules are N2(g)? if 78.09% (XN,-

Solution

3

a) partial pressure of N2 = Xn2*PtOtal

              = 0.7809*689

              = 538.04 mmhg

b)   Kh = C/Pressure(P)

K = 6.8*10^-4 M/atm

C = ?

P = 538.04/760 = 0.708 atm

(6.8*10^-4) = C/0.708

C = 4.814*10^-4 M

amount of N2 dissolved in 2 L NH3 = M*V*Mwt of N2

                                 = 4.814*10^-4*2*28

          = 0.027 g