Question

State Henry's law calculate the solubility of co2 in water at 298 Kelvin under 760 mm Hg pressure.
( kH for seo2 in water at 298 kelvin is 1.25 into 10 to the power 6 mm Hg)

Answer 1

The solubility of CO2 in water is 0.033 mol/litre

Pressure= 760 mmhg (Given)

Temp = 298 Kelvin (Given)

According to Henry's Law - Kh × x

= 760 / (1.25 × 10⁶)

= 608 × 10`6

= 6.08 × 10`4

Thus, solubility in terms of mole fraction =  6.08 × 10`4

In terms of mol / litre, solubility will be -

where, x = number of moles of solute/ total no. of moles in solution

x = number of moles of solute/ ( number of moles of solute + number of moles of water in 1 litre)

=  x = n / (n+55.55)

As, moles of CO₂ < moles of water

x = n / 55.55

= 6.08 × 10`4 = n / 55.55

= n = 6.08 × 10⁻⁴ × 55.55

n = 6.08 × 10⁻⁴ × 55.55

n = 337.74 × 10⁻⁴

n = 0.033

Thus, solubility = 0.033 mol/litre

Answer 2

Answer:

Henry's Law:  Henry's law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas.

Solubility $\propto$ Pressure of the gas

OR

The fraction of a gas in its solution in a solvent is directly proportional to the partial pressure of the gas over the solution.

OR

The partial pressure of the gas in the vapor phase is directly proportional to the mole fraction of the gas in the solution.

$p = K_H \cdot \chi$

where,

$p = partial \hspace{1mm} pressure\hspace{1mm} of\hspace{1mm} the\hspace{1mm} gas\\K_H = Proportionality \hspace{1mm} constant\\\chi = Mole \hspace{1mm} fraction \hspace{1mm}$

Given:

Temperature, T = 298 K

Pressure of CO₂ gas, p = 760 mm of Hg

$K_H$ = 1.25 × 10⁶ mm of Hg

Find:

Solubility of the gas in water.

Solution:

Let the mole fraction of the gas be $\chi$.

Now, using the formula we have,

$p = K_H \cdot \chi$

$\chi = \frac{p}{K_H}$

$\chi = \frac{760}{1.25 \times 10^6} \\\chi = 608 \times 10^{-6}\\\chi = 6.08\times 10^{-4}$

Mole fraction = Number of moles of solute/ Total number of moles of solution

$\chi = \frac{Number\ of \ moles \ of \ CO_2 }{Number \ of \ moles \ of \ solute + Number \ of \ moles \ of \ water \ in \ 1 \ litre}$

Let the number of moles of CO₂ be n.

$\therefore \chi = \frac{n}{n + 55.5}$

But n <<<< 55.5

∴ We neglect n in the denominator.

Now, we have

$\chi = \frac{n}{55.55} \\n = \chi (55.55)\\n = 6.08\times 10^{-4} \times 55.55\\n = 0.0338$

Hence, Solubility = 0.0338 moles per litre.

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