Question

10 ml of oxygen is subjected to silent electric discharge. The final volume is 8 ml. Mole fraction of ozone in the mixture is​

Answer 1

Answer:

87 mL and 81 mL

Explanation:

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Answer 2

Given:

10 ml of oxygen is subjected to silent electric discharge

To find:

The mole fraction of ozone in mixture.

Explanation:

The initial volume of $O_{2}$=$10ml$

$%of O_{2}=90%$$volume of O_{2} converted =90$%$10ml$

$=\frac{90}{100}*10$

$=0.9*10\\=9ml$$volume of O_{2} unit=(10-9)ml\\=1ml$

$3O_{2}$→$2O_{3}$

$3v$→$2v$

$v$→$\frac{2v}{3}$

$9ml$→$\frac{2}{3}*9ml$

$6ml$$V_O_{2} +V_O_{3}=(1+6)ml=7ml$

$=(7-6)ml=1ml$

The 7ml and 1ml.