Physical Chemistry> Mole Concept...

600 mL of a mixture of Ozone(O3) and Oxygen(O2) weighs 1 g at NTP.Calculate the volume of Ozone(O3)in the mixture.

Anik Chatterjee , 13 Years ago

Grade 11

5 Answers

Harish R

Last Activity: 13 Years ago

Assume xg of Ozone . O2= 1-x g

(Total moles of gas)/22.4 = 0.6 L

{x/48 + (1-x)/32} / 22.4 = 0.6

Solve for x...

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Approved

Paras Sharma

Last Activity: 8 Years ago

Hey x will give the amount of O3 to find the volume of O3 we need to apply that V/22.4 = no. of moles = x/48

Here x=3/7 g

therefore :-

V=0.2 Lt.

AVINASH mandre

Last Activity: 8 Years ago

.896/22.4=x/44 +(1.28-x)/28 .04=(28x+56.32-44x)/123216x=56.3-49.28x=0.44 gbasically we are comparing molesby Avinash mandre

Hari Krishnan

Last Activity: 7 Years ago

32g of O₂= 22.4l

1g=22.4/32 l

xg= 22.4x/32

Similarly, (1-x)g O3=(1-x)22.4/48 litres

Sum=> [22.4/x +(1-x)22.4/48] litres = 0.6 litres

Solving, x= 4/7 g Oxygen

(1-x)= 3/7 g Ozone

ie. 3/7 ×22.4/48 litres = 0.2 litres

=200 ml

Gopi Yadav

Last Activity: 6 Years ago

Let the volume of ozone be V ml.
Volume of oxegen = (600-V) ml
Mass of 22,400 ml of ozone at STP = 48 g (Molar mass of O3 is 48 )
Mass of V ml of ozone = (48 x V/22400) g
Mass of 22,400 ml of oxygen = 32 g (Molar mass of O2 is 32)
Mass of (600-V) ml of oxygen = [32 x (600-V)}g]/22400
Now, [ 48 x V/22400 + 32 x (600-V)/22400]g = 1 g
[48 x V + 32 x (600-v)]/22400 = 1

48 x V + 32 x 600 - 32 V = 22400
48 V- 32 V = 22400 - (32 x 600)
16 V = 3200
V = 200 ml.