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Byju's Answer

Standard XII

Chemistry

Depression in Freezing Point

1.The Henry's...

Question

1.The Henry's law constant for the solubility of N2gas in water at 298k is 10^5 atm.The mole fraction of N2 is 0.8.Find the number of N2 from air dissolved in 10 moles of water at 298K and 5 atm pressure

2.How many moles in 1litre of water

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Solution

P(n2) = Kh * X(n2)

rearrange the formula

X(n2) = P(n2) / Kh

= 0.8 x 5 atm / 1 x 10^5 atm
= 4 x 10^-5

now 1 mol of water has 4 x 10^-5 moles of N2 dissolved from air
so 10 moles of water has 10 x 4 x 10^-5 = 4 x 10^-4 moles of N2 dossolved from air

so the answer is 4 x 10^-4 moles of N2

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1.The Henry's law constant for the solubility of N2gas in water at 298k is 10^5 atm.The mole fraction of N2 is 0.8.Find the number of N2 from air dissolved in 10 moles of water at 298K and 5 atm pressure 2.How many moles in 1litre of water

P(n2) = Kh * X(n2) rearrange the formula X(n2) = P(n2) / Kh = 0.8 x 5 atm / 1 x 10^5 atm = 4 x 10^-5 now 1 mol of water has 4 x 10^-5 moles of N2 dissolved from air so 10 moles of water has 10 x 4 x 10^-5 = 4 x 10^-4 moles of N2 dossolved from air so the answer is 4 x 10^-4 moles of N2

1.The Henry's law constant for the solubility of N2gas in water at 298k is 10^5 atm.The mole fraction of N2 is 0.8.Find the number of N2 from air dissolved in 10 moles of water at 298K and 5 atm pressure

2.How many moles in 1litre of water

P(n2) = Kh * X(n2)

rearrange the formula

X(n2) = P(n2) / Kh

= 0.8 x 5 atm / 1 x 10^5 atm
= 4 x 10^-5

now 1 mol of water has 4 x 10^-5 moles of N2 dissolved from air
so 10 moles of water has 10 x 4 x 10^-5 = 4 x 10^-4 moles of N2 dossolved from air

so the answer is 4 x 10^-4 moles of N2