If O_2 gas is bubbled through water at 303 ; | vedclass

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

Question

(B) According to Henry's Law,$p = K_H 	imes x$,where $x$ is the mole fraction of the gas in the solution.Given $p = 0.920 \; bar$ and $K_H = 46.82 \;

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) According to Henry's Law,$p = K_H 	imes x$,where $x$ is the mole fraction of the gas in the solution.Given $p = 0.920 \; bar$ and $K_H = 46.82 \; kbar = 46820 \; bar$.$x = \frac{p}{K_H} = \frac{0.920}{46820} \approx 1.965 	imes 10^{-5}$.Since the solubility is very small,the number of moles of $H_2O$ in $1 \; L$ $(1000 \; g)$ is $n_{H_2O} = \frac{1000}{18} \approx 55.55 \; mol$.$x = \frac{n_{O_2}}{n_{O_2} + n_{H_2O}} \approx \frac{n_{O_2}}{n_{H_2O}}$.$n_{O_2} = x 	imes n_{H_2O} = (1.965 	imes 10^{-5}) 	imes 55.55 \approx 1.09 	imes 10^{-3} \; mol$.$n_{O_2} \approx 1.09 \; millimoles$.The nearest integer is $1$.

Similar Questions

Which among the following salts,solubility decreases with increase in temperature?

Assertion $(A)$: For an endothermic dissolution process,an increase in temperature increases the solubility in a nearly saturated solution.
Reason $(R)$: In a saturated solution,dynamic equilibrium exists between the dissolved solute and the undissolved solute.

Calculate the solubility of a gas in water at $260 \ mm \ Hg$ and $25^{\circ} C$ if Henry's law constant of the gas is $0.159 \ mol \ dm^{-3} \ atm^{-1}$ at $25^{\circ} C$.

Assertion : The molecules of the dissolved gas present in a liquid gain kinetic energy as temperature is raised.
Reason : Gases tend to be more soluble in liquids as the temperature is raised.

Maximum amount of a solid solute that can be dissolved in a specified amount of a given liquid solvent does not depend upon $ . . . . . . $ .
$(i)$ Temperature $(ii)$ Nature of solute $(iii)$ Pressure $(iv)$ Nature of Solvent

Vedclass Test Series

Exam Paper Generator

Online Exam Module