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Question: When a gas is bubbled through water at 298 K, a very dilute solution of gas is obtained. Henry’s law...

When a gas is bubbled through water at 298 K, a very dilute solution of gas is obtained. Henry’s law constant for the gas is 100 kbar. If gas exerts a pressure of 1 bar, the number of moles of gas dissolved in 1 litre of water is

A

0.5550.555

B

55.55×10−555.55 \times 10 ^ { - 5 }

C

55.55×10−355.55 \times 10 ^ { - 3 }

D

5.55×10−55.55 \times 10 ^ { - 5 }

Answer

55.55×10−555.55 \times 10 ^ { - 5 }

Explanation

Solution

p=kH×x\mathrm { p } = \mathrm { k } _ { \mathrm { H } } \times \mathrm { x }

Mole fraction = Moles of gas  Total moles = \frac { \text { Moles of gas } } { \text { Total moles } }

Moles of H2O=100018=55.55(∵1 L=1000 g)\mathrm { H } _ { 2 } \mathrm { O } = \frac { 1000 } { 18 } = 55.55 \quad ( \because 1 \mathrm {~L} = 1000 \mathrm {~g} )

Mole fraction =xx+55.55(55.55>>x)= \frac { x } { x + 55.55 } ( 55.55 > > x )