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Standard XII

Chemistry

Gases in Liquids

Compute the m...

Question

Compute the molar solubility of water that is saturated with air.

[Given: $k =$ 0.00105 M/atm, $P_{N e}$ in air = 0.0341 atm]

2.56 \times 10^{- 2} M

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3.58 \times 10^{- 5} M

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4.44 \times 10^{- 6} M

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2.78 \times 10^{- 4} M

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Solution

The correct option is B $3.58 \times 10^{- 5} M$
The expression for Henry's law is $C = k P_{g a s}$

Substitute values in the above expression:

$C = (0.00105 M / a t m) (0.0341 a t m) = 3.58 \times 10^{- 5} M$

The molar solubility in water (saturated with air) is $3.58 \times 10^{- 5} M$

Hence, the correct answer is option $B$ .

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Compute the molar solubility of water that is saturated with air.[Given: k= 0.00105 M/atm, PNe in air = 0.0341 atm]

The correct option is B 3.58×10−5MThe expression for Henry's law is C=kPgasSubstitute values in the above expression:C=(0.00105M/atm)(0.0341atm)=3.58×10−5MThe molar solubility in water (saturated with air) is 3.58×10−5MHence, the correct answer is option B.

Compute the molar solubility of water that is saturated with air.

[Given: $k =$ 0.00105 M/atm, $P_{N e}$ in air = 0.0341 atm]