How many molecules of CO2 gas is passed in 900ml water at 298 K temperature - The value of KH is 6-02 - 10-4 bar and partial pressure of CO2 gas is 2 - 10-8 bar-

According to Henry's law, p_CO_2 = K_H.χ p_CO_2 = partial pressure of CO_2 = 2×10^ 8 bar K_H = Henry's law constant = 6.02×10^ 4bar χ = mo ...

188

You visited us 188 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Ideal Gas Equation

How many mole...

Question

How many molecules of $C O_{2}$ gas is passed in $900$ ml water at $298$ K temperature . The value of $K_{H}$ is $6.02 \times 10^{- 4}$ bar and partial pressure of $C O_{2}$ gas is $2 \times 10^{- 8} b a r$ :

Open in App

Solution

According to Henry's law, $p_{C O_{2}} = K_{H} . χ$
$p_{C O_{2}} = partial pressure of C O_{2} = 2 \times 10^{- 8} b a r$
$K_{H} = Henry's law constant = 6.02 \times 10^{- 4} b a r$
$χ=mole fraction of CO2$
$χ = \frac{number of moles of C O_{2}}{number of moles of C O_{2} + number of moles of H_{2} O} = \frac{number of moles of C O_{2}}{number of moles of H_{2} O}$
(since, the moles of $C O_{2}$ are negligible in comparison to $H_{2} O$ )
number of moles of $H_{2} O = \frac{given weight of H_{2} O}{molecular weight of H_{2} O} = \frac{900 g}{18 g} = 50$
(As, $1 m l = 1 g$ so, $900 m l = 900 g$ )
$Number of moles of C O_{2} = \frac{p_{C O_{2}} \times number of moles of H_{2} O}{K_{H}}$
$Number of moles of C O_{2} = \frac{2 \times 10^{- 8} \times 50}{6.02 \times 10^{- 4}} = \frac{10^{- 2}}{6.02}$
$Number of molecules of C O_{2} = \frac{10^{- 2}}{6.02} \times 6.02 \times 10^{23} = 10^{21} m o l e c u l e s$

Suggest Corrections

Similar questions

^{N₂gas is bubbled through water at 298K. How many moles of N₂ gas would dissolve in 1L water assumed that partial pressure of gas is = 0.987 bar and K_H = 76.48 K .}

Q. How many oxygen is dissolved in 100 mL water at 298 K if partial pressure of oxygen is 0.5 atm and

K_{H} = 1.4 \times 10^{- 3}

mol/L/atm?

Q. How many grams of

C O_{2}

gas is dissolved in a

1 l i t r e

bottle of carbonated water if the manufacturer uses a pressure of 2.4 atmosphere in the bottling process at

25^{o} C ? (

given

K_{H}

C O_{2}

water

= 9.6 a t m / m o l / l

25^{o} C

)

Q. How many grams of

{C O}_{2}

gas is dissolved in a

1

lt bottle of carbonated water if the manufacturer uses a pressure of

2.4

atmosphere in the bottling process at

25^{o} C

K_{H}

{C O}_{2}

water

= 29.76 a t m / m o l e / l

25^{o} C

)

Henrys law constant for $C O_{2}$ in water is $1.6 \times 10^{8}$ Pa at 298K.

Calculate quantity of $C O_{2}$ in 500 mL of soda water when packed under 2.5 atm $C O_{2}$ pressure at 298K.

Join BYJU'S Learning Program

How many molecules of CO2 gas is passed in 900ml water at 298 K temperature . The value of KH is 6.02×10−4 bar and partial pressure of CO2 gas is 2×10−8bar:

How many molecules of $C O_{2}$ gas is passed in $900$ ml water at $298$ K temperature . The value of $K_{H}$ is $6.02 \times 10^{- 4}$ bar and partial pressure of $C O_{2}$ gas is $2 \times 10^{- 8} b a r$ :