If N_2 gas is bubbled through water at 293 K | vedclass

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(C) According to Henry's law: $P = K_{H} X$ First,calculate the mole fraction of $N_2$ $(X_{N_2})$: $X_{N_2} = \frac{P_{N_2}}{K_{H}} = \frac{0.987 \ b

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$7.16 	imes 10^{-4}$

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(C) According to Henry's law: $P = K_{H} X$ First,calculate the mole fraction of $N_2$ $(X_{N_2})$: $X_{N_2} = \frac{P_{N_2}}{K_{H}} = \frac{0.987 \ bar}{76.48 	imes 10^3 \ bar} = 1.29 	imes 10^{-5}$ Calculate the number of moles of water in $1 \ litre$ $(1000 \ g)$: $n_{H_2O} = \frac{1000 \ g}{18 \ g/mol} = 55.5 \ mol$ Since $n_{N_2} \ll n_{H_2O}$,we use the approximation $X_{N_2} \approx \frac{n_{N_2}}{n_{H_2O}}$: $n_{N_2} = X_{N_2} 	imes n_{H_2O} = 1.29 	imes 10^{-5} 	imes 55.5 \ mol = 7.16 	imes 10^{-4} \ mol$

If $N_2$ gas is bubbled through water at $293 \ K$,how many moles of $N_2$ gas would dissolve in $1 \ litre$ of water? Assume that $N_2$ exerts a partial pressure of $0.987 \ bar$. [Given $K_{H}$ for $N_2$ at $293 \ K$ is $76.48 \ kbar$]

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If $N_2$ gas is bubbled through water at $293 \ K$,how many millimoles of $N_2$ gas would dissolve in $1 \ L$ of water? Assume that $N_2$ exerts a partial pressure of $0.987 \ bar$. Given that Henry's law constant for $N_2$ at $293 \ K$ is $76.48 \ kbar$.

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