Question

672 ml of ozonised oxugen weighs 1 g. calculate the volume of ozone in ozonised oxygen

Answer 1

Let us consider the volume of O2 (oxygen) = x mL

So, the volume of O3 (ozone) = 672 - x mL

Since, Mass of (O2 + O3) = 672 mL

Thus, Weight of x mL of O2 =
32 × x
22400 gm
[because, molar mass of O2 = 2 x 16 = 32 g/mol]---- eq 1

Similarly, Weight of 672 - x mL of O3 =
48 × (672 − x)
22400 gm
[because, molar mass of O3 = 3 x 16 = 48 g/mol]--- eq 2

Here, 22400 gm or 22400 mL is the standard molar volume of O at NTP

​Now, equating eq 1 and 2, we get

32 × x. +. 48 × (672 − x)
22400 gm. 22400 gm

= 1
So, weight of x mL of O = x moles x atomic weight

32x + 32256 - 48x = 22400

32256 - 22400 = 48 x - 32 x

9856 = 16 x

So, x =
9856
16
= 616 mL

​So, the volume of oxygen in ozonized oxygen = 616 mL

Volume of ozone in the ozonized oxygen = (672 - 616) mL = 56 mL

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Answer 2

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