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 ...

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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 $2 \times 10^{- 8} b a r$ :

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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, $900 m l = 900 g$ )
$Number of moles of C O_{2} = \frac{p_{C O_{2}} \times number of moles of H_{2} O}{K_{H}}$
$Number of moles of C O_{2} = \frac{2 \times 10^{- 8} \times 50}{6.02 \times 10^{- 4}} = \frac{10^{- 2}}{6.02}$
$Number of molecules of C O_{2} = \frac{10^{- 2}}{6.02} \times 6.02 \times 10^{23} = 10^{21} m o l e c u l e s$

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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 $2 \times 10^{- 8} b a r$ :