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Byju's Answer

Standard IX

Chemistry

Ways of Representing the Concentration of a Solution

At 25∘C and...

Question

At $25^{\circ} C$ and $1$ atm pressure, the mole fraction of $N_{2}$ in air is $0.90$ . Calculate the concentration of $N_{2}$ present in a glass of water at C. (Given $K_{H}$ for $N_{2}$ at $25^{\circ} C$ is $8.5 \times 10^{- 7}$ M/mm Hg)

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Solution

Given $P_{N_{2}} = 1 a t m$
$P_{N_{2}} = K_{H} \times N_{2}$
From Henry's law,
$X_{N_{2}} = \frac{P_{N_{2}}}{K_{H}} = \frac{1}{8.5} \times 10^{- 7}$
$\Rightarrow X_{N_{2}} = \frac{n (N_{2})}{n (N_{2}) + n (H_{2} O)} \Rightarrow \frac{n (N_{2})}{n (H_{2} O)}$
$\Rightarrow n (H_{2}) = X_{N_{2}} . n (H_{2} O)$
$= \frac{1}{8.5 \times 10^{- 7}} \times 55.5$
Concentration of $N_{2} = 6.53 \times 10^{7}$ mol/lit

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At 25∘C and 1 atm pressure, the mole fraction of N2 in air is 0.90. Calculate the concentration of N2 present in a glass of water at C. (Given KH for N2 at 25∘C is 8.5×10−7 M/mm Hg)

Given PN2=1atmPN2=KH×N2From Henry's law,XN2=PN2KH=18.5×10−7⇒XN2=n(N2)n(N2)+n(H2O)⇒n(N2)n(H2O)⇒n(H2)=XN2.n(H2O) =18.5×10−7×55.5Concentration of N2=6.53×107 mol/lit

At $25^{\circ} C$ and $1$ atm pressure, the mole fraction of $N_{2}$ in air is $0.90$ . Calculate the concentration of $N_{2}$ present in a glass of water at C. (Given $K_{H}$ for $N_{2}$ at $25^{\circ} C$ is $8.5 \times 10^{- 7}$ M/mm Hg)