Question

calculate the number of molecules of oxygen gas that occupies a volume of 224 ml at 273k and 3atm pressure ​

Answer 1

Answer:

The Answer is

Explanation:

22400 ml. of oxygen at 273 K temperature and 3-atm pressure = 22400 ×3/1= 67200 ml at STP.

So 224 ml of oxygen at 273 K and 3-atm = 224 ×3= 672 ml at STP

The number of Oxygen molecules present in 22400 ml of gas at STP=N

(where N= Avagadro Number)

Therefore 224 ml of oxygen at 273K and 3-atm pressure will contain =N×672/22400 × N molecules.

=3/100×6.022×10

23

=18.066×10

21

Molecules =1.8066×10

22

molecules

Answer 2

Answer:

Brainly.in

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Gokul0Indhalur

Gokul0Indhalur

15.06.2018

Chemistry

Secondary School

+5 pts

Answered

calculate the number of molecules of oxygen gas that occupies a volume of 224 ml at 273 K and 3 atm pressure.

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NiharikaT

NiharikaT

From the Question,

Volume of the Gas(V) = 224 mL.

= 0.224 liter.

Temperature of the Gas(T) = 273 K.

Pressure of the Gas(T) = 2 atm

Now, Using the Formula,

PV = nRT

where, R = Gas Constant = 0.082 L-atm mol⁻¹ K⁻¹

and n = number of moles.

∴ 2 × 0.224 = n × 0.082 × 273

∴ 0.448 = 22.386 × n

⇒ n = 0.448/22.386

⇒ n = 0.02 moles.

Now,

Using the Formula,

No. of molecules = No. of moles × 6.022 × 10²³

= 0.02 × 6.022 × 10²³

= 0.121 × 10²³

Thus, the number of the molecules of the Oxygen gas is 0.121 × 10