Question

Calculate the mole fraction of CO2 in 1l of soda water sealed under pressure of 3.5 bar at 298k.kh is equal to 1.67×10^3bar

Answer 1

Answer:

Explanation:

It is given that:

KH= 1.67 × 108Pa

PCO2 = 2.5 atm = 2.5 × 1.01325 × 105Pa

= 2.533125 × 105Pa

 

According to Henry's law:

PCO2 = KHX

⇒ x =  PCO2 / KH

= 2.533125 × 105 / 1.67 × 108

= 0.00152

 

We can write,  

[Since, is negligible as compared to]

 

In 500 mL of soda water, the volume of water = 500 mL

[Neglecting the amount of soda present]

We can write:

500 mL of water = 500 g of water

=500 / 18 mole of water

= 27.78 mol of water

Now, nCO2 / nH2O = x

nCO2 / 27.78 = 0.00152

nCO2 = 0.042 mol

Hence, quantity of CO2 in 500 mL of soda water = (0.042 × 44)g

= 1.848 g

hope it works

Answer 2

Answer:

The mole fraction of $CO_2$ is $2.095\times 10^{-3}$.

Explanation:

It is given in the question that,

The volume of soda water is 1L.

The pressure is 3.5 bar at the temperature 298K.

And, $K_H = 1.67\times10^3bar$

now, According to the Henry's Law,

$P=K_H\times x$

Substitute the given values in the following equation,

$3.5=1.67\times10^{3}bar\times x$

Determine the value of x by using calculator,

$x=\frac{3.5}{1.67\times10^3} \\x=2.095\times 10^{-3}$

The mole fraction of $CO_2$ is $2.095\times 10^{-3}$.

To know more about the mole fraction, click on the link below,

https://brainly.com/question/14783710

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