1.

N_(2) exerts a pressure of 0.987 bar. The mole fraction of N_(2) ia (K_(H)=74.48 k bar)

Answer»

`7.648 xx 10^(-3)`
`9.87 xx 10^(-5)`
`1.3 xx 10^(-5)`
`2.6 xx 10^(-5)`

Solution :`P_(N2)=K_(H) xx X_(N2)`
`:. X_(N2)=P_(N2)/K_(H)=(0.987"barr")/(76480"bar")=1.29 xx 10^(-5)`


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