$\mathrm{CO}_{2}$ gas is bubbled through water during a soft drink manufacturing process at 298 K . If $\mathrm{CO}_{2}$ exerts a partial pressure of 0.835 bar then $\times \mathrm{m} \mathrm{mol}$ of $\mathrm{CO}_{2}$ would dissolve in 0.9 L of water. The value of $x$ is (Nearest integer) (Henry's law constant for $\mathrm{CO}_{2}$ at 298 K is $1.67 \times 10^{3}$ bar)

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Solutions

CO_{2}

gas is bubbled through water during a soft drink manufacturing process at 298 K . If

CO_{2}

exerts a partial pressure of 0.835 bar then

\times m mol

CO_{2}

would dissolve in 0.9 L of water. The value of

x

is (Nearest integer)
(Henry's law constant for

CO_{2}

at 298 K is

1.67 \times 1 0^{3}

bar)

Answer: 25

Solution

Henry's law relates the partial pressure of the gas to its mole fraction in solution,

p = K_{H} x

, so

x = \frac{0.835}{1.67 \times 1 0 ^{3}} = 5.0 \times 1 0^{- 4}

Taking the density of water as

1 g m L^{- 1}

, 0.9 L contains

\frac{900}{18} = 50

mol. The dissolved gas is a trace, so the mole fraction is effectively

n_{C O_{2}} / n_{H_{2} O}

and

n_{C O_{2}} = 5.0 \times 1 0^{- 4} \times 50 = 2.5 \times 1 0^{- 2} mol

The amount of carbon dioxide that dissolves is

25

millimole.

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Integer answer

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