672 ml of ozonised oxygen at NTP were found to be weighing 1g .Calculate volume of ozone in the ozonised oxygen..
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Answered by
3
Hello! ^ ^ Let me try to answer this question :)
please remember some formula:
- P V = n R T
- n = m ÷ Mr
P = Pressure (atm)
V = Volume (litre, L)
n = Mole (mole)
R = 0.082 L atm/mole K
T = Temperature (Kelvin, K)
m = mass (gram)
Mr = molecular mass (gram/mole)
Thank you, please correct if there are any mistakes ^ ^
please remember some formula:
- P V = n R T
- n = m ÷ Mr
P = Pressure (atm)
V = Volume (litre, L)
n = Mole (mole)
R = 0.082 L atm/mole K
T = Temperature (Kelvin, K)
m = mass (gram)
Mr = molecular mass (gram/mole)
Thank you, please correct if there are any mistakes ^ ^
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Answered by
0
Answer:
Let, the volume of O
3
=x mL.
Volume of O
2
present =(600−x) mL
22400 mL of O
3
and O
2
at STP will weight 48 g and 32 g respectively.
The weight of x mL of O
3
=
22400
x×48
g.
The weight of (600−x) mL of O
2
=
22400
(600−x)
×32.
Total weight of ozonised O
2
(600mL) =
22400
48x
+
22400
(600−x)×32
=1.0.
∴x=200 mL
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