Question

Nitrogen gas is dissolved in water at STP with partial pressure 7.425 torr calculate the moles and milli moles present in th solution

Answer 1

hello!

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According to Henry's law,

According to Henry's law,p N 2

According to Henry's law,p N 2

According to Henry's law,p N 2

According to Henry's law,p N 2 =K H

According to Henry's law,p N 2 =K H

According to Henry's law,p N 2 =K H .X N 2

According to Henry's law,p N 2 =K H .X N 2

According to Henry's law,p N 2 =K H .X N 2

According to Henry's law,p N 2 =K H .X N 2

According to Henry's law,p N 2 =K H .X N 2 ⇒x N 2

According to Henry's law,p N 2 =K H .X N 2 ⇒x N 2

According to Henry's law,p N 2 =K H .X N 2 ⇒x N 2

According to Henry's law,p N 2 =K H .X N 2 ⇒x N 2

According to Henry's law,p N 2 =K H .X N 2 ⇒x N 2 = 76800 =0.987

0.987

0.987 =1.29×10 −5

−5

−5 Moles of water in 1L= 18 = 1000

1000

1000 =55.56

1000 =55.56Let us take the moles of nitrogen as n

1000 =55.56Let us take the moles of nitrogen as nTotal moles =n+55.56

1000 =55.56Let us take the moles of nitrogen as nTotal moles =n+55.56∴ x N 2

1000 =55.56Let us take the moles of nitrogen as nTotal moles =n+55.56∴ x N 2

1000 =55.56Let us take the moles of nitrogen as nTotal moles =n+55.56∴ x N 2

1000 =55.56Let us take the moles of nitrogen as nTotal moles =n+55.56∴ x N 2

1000 =55.56Let us take the moles of nitrogen as nTotal moles =n+55.56∴ x N 2 = n+55.56 n

n

n =1.29×10 −5

−5

−5 ⇒n(1−1.29×10 −5

−55.56×1.29×10 −5 =0

=0⇒n=0.7×10 −3

=0.7 milimoles