Solution: We know, Henry law is P = K × X given P = 0 987 bar K = 76 48 ×103 bar X = 0 98776480 = 0 0000129 = 1 29 ×10-5 (Where X is mole fraction of N2) No of moles of water = Given massmolar mass = 1000 ml×1g/ml18g/mol= 55 55 moles Total moles = moles of nitrogen + moles of water = n+ 55 55 X = nn+55 55=1 29 ×10-5 n = 0 7×10-3 = 0 7 millimoles

Plz solve this.....

Solution:
We know,
Henry law is P = K

\times

X
given P = 0.987 bar
K = 76.48

\times 10^{3}

bar
X =

\frac{0.987}{76480}

= 0.0000129 =

1.29 \times 10^{- 5}

(Where X is mole fraction of N₂)
No. of moles of water =

\frac{G i v e n m a s s}{m o l a r m a s s} = \frac{1000 m l \times 1 g / m l}{18 g / m o l}

= 55.55 moles
Total moles = moles of nitrogen + moles of water = n+ 55.55
X =

\frac{n}{n + 55.55}

1.29 \times 10^{- 5}

n =

0.7 \times 10^{- 3}

= 0.7 millimoles

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