Chemistry, asked by Anonymous, 1 year ago


calculate \: the \: number \: of \: molecules \: present \: in  \\ 11.2 \: litres \: of \: oxygen \: at \:normal \: temperature \: and  \\ pressure.


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Answers

Answered by Rathin5678
3

hope this answer helps you and plz follow me. mark it as the Brainliest plz

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Answered by SunTheHelpingHand
3

I hope u already got the answer from another user..

So...my work is nothing but to explain the problem in a simple theoretical way rather than adopting a 'typical problem solving method'

So...lemme explain what is NTP?

➡️NTP(Normal Temperature and Pressure)

It is the condition for a gas existing at Normal Temperature and Pressure.

So what's Normal Temperature and Pressure?

Here we take Normal Temperature as 20°C/293K(most places have an average temperature of 20°C)

and pressure 1atm/101.325kPa(our atmospheric pressure).

So NTP is ideal for a lab condition.

Suppose u do any experiment in ur Kuwait school lab.So all u r doing it under NTP(under normal temperature and pressure).

so...'ya' familiar with NTP...XD

So let's have a look on STP...

Yes Mr. STP.

So 'he' stands for Standard Temperature and Pressure.

He is just like a school Principal.

All his ideas are 'ideal'. But not 'applicable' in normal conditions XD XD XD...

➡️STP(Standard Temperature and Pressure):

It is the condition for a gas existing at Standard Temperature and Pressure.

So here the standard temperature is 0°C, which is not applicable in common labs!

That's why I called him Mr.Ideal..XD

Anyway, here the pressure is also different, Here pressure is 100kPa/0.987atm.

➡️So.. let's come back to our question...

At NTP, 'one' mole of any gas occupies 22.4L.

So....here we are provided with only 11.2L. of 02 gas.

As u know, one mole of gas contains 'Avogadro number' of gas molecules,

ie.. in 1 mole= 6.022×10²³ molecules,

so...in 11.2L, there's only ½ mole, so ½×(6.022×10²³)

So 3.011×10²³ molecules.

So we can find no. of molecules in any gas, or any mixture, stating it's NTP/STP conditions☺️✌️

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SunTheHelpingHand: :)
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