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`        how to calculate no. of moles?`
8 years ago

8 Points
```										The calulation depends. But in general a mole is the no of units that contain an Avagadros number of chemical units. As in, a mole of oxygen atoms => 6.023X1023   atoms of oxygen . The number of moles of a chemical species present in a sample is obtained by various methods.
The easiest method is by knowing the mass of the chemical species present in a sample. For example. in a vessel you have 10 kg of oxygen.
We know that in that sample oxygen exists as a gas and hence oxygen molecules. We know ( these are standard values and must be remembered) oxygen has a molecular mass of 32g/mol. What this implies is that if you take 6.023X10^23 molecules of oxygen, then they will always weigh 32 g. Therefor one mole of oxygen gas weighs 32 g. You can easily find out how many moles are there in 10kg by unitary method.(10,000/32 unit:mol).
Depending on situation and qsn given you might have to use stoichiometry of a reaction or gas laws to obtain the same too.
```
8 years ago
Ramesh V
70 Points
```										For calculating the no. of moles of molecules or atoms
Suppose u have a sample of weight 'w' in gms which is chemically homogeneous

Now
1. No of moles of atoms = w/ atomic weight

(If your sample is single element like 02 ,N2 then u can use this formula
Ex- suppose a sample of oxygen have 32 gms
Then no of moles of atoms of Oxy. = 32/16 = 2 moles
In other words this tells you that ur sample contains 2 moles of Oxy. atoms or 2 x 6.023 x 10 ^23 no. of Oxy. atoms

2. No. of moles of molecules= w / molecular weight

If your sample is single element or compound like NaCl, AgNO3 etc then u can use this formula
For the above example
Then no of moles of molecules of Oxy. = 32/32 =1 mole
In other words this tells you that ur sample contains 1 mole of Oxy. molecules or 6.023 x 10 ^23 no. of Oxy. molecules
Ex-2 117 gms of common salt (NaCl or Sodium Chloride)
No of moles of molecules = 117 / (23 + 35.5) = 2 moles
--
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we will get you the answer and detailed solution very quickly.We are all IITians and here to help you in your IIT JEE preparation. All the best.
Regards,
Naga Ramesh
IIT Kgp - 2005 batch
```
8 years ago
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