Henry's law constant for $C O_{2}$ in water is $1.67 \times 10^{8} p a$ at $298 K$ the quantity of $C O_{2}$ in $500 m l$ of soda water when packed under $2.5 a t m$ pressure is:

0.082 m o l e

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0.82 m o l e

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0.41 m m o l e

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0.042 m o l e

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Solution

The correct option is B $0.042 m o l e$
$K_{x} = 1.67 \times 10^{8} P a$

According to Henry's law,
$P = K_{x} \times X$

$X = \frac{2.5 \times 1.01325 \times 10^{5}}{1.67 \times 10^{8}}$

$= 1.52 \times 10^{- 3}$

Since we have $500$ soda water

$∴$ Volume of water $= 500 m l$

No. of moles $= 27.78$

$\Rightarrow$ Mole fraction = $n_{C O_{2}} / 27.78$

$n_{C O_{2}} = 0.042$

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Henry’s law constant for carbon dioxide in water is 1.67 × 108 Pa at 298K. Calculate the quantity
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Q. The quantity of

{C O}_{2}

in 500ml of soda water when packed under 2.5 atm

{C O}_{2}

pressure at 298K is ____ gm ( Henry's law constant

1.67 \times 10^{8}

pa at 298K)

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C O_{2}

in water is

1.678

10^{8}

pa at 298K the quantity of

C O_{2}

in 500ml of soda water when packed under 2.5 atm pressure is :

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.

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Henry's law constant for CO2 in water is 1.67×108pa at 298 K the quantity of CO2 in 500 ml of soda water when packed under 2.5 atm pressure is:

The correct option is B 0.042 moleKx=1.67×108Pa According to Henry's law, P=Kx×X X=2.5×1.01325×1051.67×108 =1.52×10−3 Since we have 500 soda water ∴ Volume of water =500ml No. of moles =27.78 ⇒ Mole fraction = nCO2/27.78 nCO2=0.042

Henry's law constant for $C O_{2}$ in water is $1.67 \times 10^{8} p a$ at $298 K$ the quantity of $C O_{2}$ in $500 m l$ of soda water when packed under $2.5 a t m$ pressure is: