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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$7.16 	imes 10^{-4}$

Vedclass · Answer

(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 $76.48 \ kbar$]

Similar Questions

What is the solubility of a gas in water at $25^{\circ} C$ if the partial pressure is $0.18 \ atm$? $(K_{H} = 0.16 \ mol \ dm^{-3} \ atm^{-1})$

Calculate the solubility of a gas having partial pressure $0.15 \ bar$ at $25^{\circ} C$.
$[K_{H} = 0.16 \ mol \ dm^{-3} \ bar^{-1}]$

In which of the following salts,the solubility increases appreciably with an increase in temperature?

$(a)$ Explain the following phenomena with the help of Henry’s law.
$(i)$ Painful condition known as bends.
$(ii)$ Feeling of weakness and discomfort in breathing at high altitude.
$(b)$ Why does a soda water bottle kept at room temperature fizz on opening?

The solubility of a gas in a liquid generally increases with :-

Vedclass Test Series

Exam Paper Generator

Online Exam Module