Question

The following table provides some information on carbon dioxide solubility in water.

Sgas (mol L−1)	Pgas (bar)	kH (mol L−1 bar−1)	T (∘C)
0.0380	1.01		20.0
0.0980			20.0
	1.01	0.0340	25.0

Part A

What is the Henry's law constant for CO2 at 20∘C?

Express your answer in mol L−1 bar−1 to four decimal places.

	mol L−1 bar−1

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Part B

What pressure is required to achieve a CO2 concentration of 0.0980 mol L−1 at 20∘C?

Express your answer to three significant figures and include the appropriate units.

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Part C

At 1.01 bar, how many moles of CO2 are released by raising the temperature of 1 litre of water from 20∘C to 25∘C?

Express your answer to four decimal places and include the appropriate units.

Answer 1

Solution :-

Part A

Given data

Solubility of gas = 0.0380 mol L-1

Pressure of gas = 1.01 bar

KH = ?

Formula to calculate the Henry’s gas constant is as follows

C= KH * p

Where c= concentration of gas (mol L-1)

P = pressure (bar )

Now lets put the values in the formula and calculate the KH

0.0380 mol L-1 = KH * 1.01 bar

0.0380 mol L-1 / 1.01 bar = KH

0.0376 mol L-1 bar-1 = KH

Therefore Henry's law constat KH = 0.0376 mol L-1 bar-1

Part B) Given data

At 20 ^oC KH= 0.0376 mol L-1 bar-1

C= 0.0980 mol L-1

P = ?

Lets put the values in the formula

C= KH*p

0.0980 mol L-1 = 0.0376 mol L-1 bar-1* p

(0.0980 mol L-1 / 0.0376 mol L-1 bar-1) = p

2.61 bar = p

Therefore pressure needed = 2.61 bar

Part C)

Given data

P = 1.01 bar

KH= 0.0340 mol L-1 bar-1

C= ?

C= KH*p

Lets put the values in the formula

C =0.0340 mol L-1 bar-1 * 1.01 bar

C =0.0343 mol L-1

At 20 ^oC solubility is 0.0380 mol L- 1 and at 25 ^oC solubility is 0.0343 mol L-1

So the number of moles of CO2 released = 0.0380 mol – 0.0343 mol

                                                                           = 0.0037 mol CO2

So the number of moles of CO2 released at 25^oC = 0.0037 mole CO2