Question

On passing 1.8l of oxygen through ozonizer gives 1.6l of a mixture of ozone and oxygen. What is the mole fraction of ozone in the resultant mixture?

Answer 1

$3O_{2} ----------> 2O_{3}$

18ml                             0

18 - x                            2/3x

Volume of mixture = 1.8 -x + 2.3x = 1.6

=> x = 0.6

Volume of $O_{3}$ = 2/3 * 0.6 = 0.4L

Mole Fraction of $O_{3}$ = 0.4/1.6 = 0.25

Hope this helps, 平和 :)

Answer 2

The mole fraction of ozone in the given resultant mixture is 0.34.

Explanation:

The equation for given reaction will be as follows.

             $3O_{2} \rightarrow 2O_{3}$

Moles of oxygen will be as follows.

      No. of moles = $\frac{\text{volume given}}{\text{molar volume}}$

                             = $\frac{1.8 L}{22.4 L}$

                             = 0.080 mol

Total no. of moles present in the mixture will be as follows.

       No. of moles = $\frac{\text{volume given}}{\text{molar volume}}$

                             = $\frac{1.6 L}{22.4 L}$

                             = 0.071 mol

Hence,      $3O_{2} \rightarrow 2O_{3}$

              0.08         0

        0.080 - 3x      2x

Therefore, net equation will be as follows.

                   0.080 - 3x + 2x = 0.071

                      - x = - 0.009

or,                     x = 0.009

Hence, we will calculate the mole fraction of ozone in the resultant mixture as follows.

               $O_{3} = \frac{2x}{0.080 - 3x}$

                          = $\frac{2 \times 0.009}{0.080 - 3 \times 0.009}$

                          = 0.34

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