Question

Henry constant for argon is 40 k bar im water,detrmine molal concentration of argon in.water when it is stored above water at 10 bar pressure

Answer 1

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Answer 2

Answer:

The molality concentration of argon in water is $m = 1.39\times10^{-2}\ mol$.

Explanation:

Given that,

Henry constant for argon = 40 k bar

According to Henry's law,

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

It is defined as:

$P= k_{H}x$

Where, P = pressure

k = Henry constant

x = mole fraction

Now, The mole fraction of the gas in solution is

$x=\dfrac{P}{k_{H}}$

$x=\dfrac{10}{40000}$

$x = 2.5\times10^{-4}$

Mole of solute$x = 2.5\times10^{-4}$

The molality is equal to the number of moles of solute divided by the mass of solvent.

The molality of argon is

As 1 liter of water contains 55.5 mol of it.

Then,if n shows the number of moles of argon in solution.

$x_{arg}=\dfrac{number\ of\ molal}{number\ of\ mol+55.5 mol}$

Here, In the denominator, number of mole <<55.5 mole

$x_{arg}=\dfrac{number\ of\ molal}{55.5\ mol}$

Number of molal $m=2.5\times10^{-4}\times55.5$

$m = 1.39\times10^{-2}\ mol$

Hence, The molality concentration of argon in water is $m = 1.39\times10^{-2}\ mol$