Please solve: 600mL of a mixture of ozone & O2 weighs 1g at NTP.Calculate the volume of ozone in the mixture.?

note that,

molecular mass of oxygen = 32

molecular mass of ozone = 48

calculations:-

let vol of O2 = x ml

hence vol of O3 = 600 - x ml

at STP,

wt of x ml O2 = 32*x/22400 gm

wt of 600 - x ml O3 = 48*(600 - x)/22400 gm

hence,

32*x/22400 + 48*(600 - x)/22400 = 1

or, 32x + 28800 - 48x = 22400

=> 16x = 28800 - 22400 = 6400

=> x = 6400/16 = 400 ml

so vol of O2 in the mixture is 400 ml. (ans)

You use the ideal gas equation: P x V = n x R x T

P = 1 atm given

V = 0.6 L given

R = .082058 L*atm)/(K*mol) constant

T = 25 C (i'm not sure, it depends in what NTP means)

You solve for n, which gives you the total number of moles of the mixture.

Then calculate moles of O2: 1g x (1 mol / 15.999 g )

subtract moles of O2 from n (total number of moles of mixture) call the calculated value n2

Then substitute n2 back in the equation: P x V = n x R x T

This time we want to solve for V

P = 1 atm given

R = .082058 L*atm)/(K*mol) constant

T = 25 C

n2 = ... calculated

this V gives you the volume of ozone in the mixture.

hope this helps : )

You want the mole fraction of O3 in the mixture. To do this divide 22400/600 mL to get the pseudo-mole weight (PMW) of an ideal gas. It will be more than 32 and less than 48. If you call "f" the mole fraction of ozone,

..... 48 f + 32 (1-f) = PMW

and 16 f = PMW-32

and f= (PMW-32)/16

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