which sample contains more molecules, 3.14L of At at 85.0 degrees Celsius and 1111mm Hg pressure or 11.07 of Cl2?
Solution
I think it is \'Ar\' not \'At\'. But what is 11.07 of Cl 2 ? There is no unit.
If it is 11.07gm then:
Let, 3.14L of Ar at 85degree(358K) Cand 1111mm Hg pressure is equvalent to X L of Ar at STP(298K and 76mm Hg).
So, we know (P 1 V 1 /T 1 )=(P 2 V 2 /T 2 )
Then, (1111*3.14)/358=(76* X )/298
or, X =38.209 L
According to Avogadros Law 22.4L gas at STP contains 6.023x10 23 no of atoms.
So, 38.209L of Ar contains (6.023x10 23 x38.209)/22.4= 1.02738x10 24 no of atoms.
For Ar no of atoms and no of molecules are same, as it is a innert gas.
Now atomic wt. of Cl 2 is 35.45
So, 35.45gm of Cl 2 contains 6.023x10 23 atoms
So for 11.07 gm of Cl 2 no of atoms= (6.023x10 23 x11.07)/35.45=1.8808x10 23 atoms.
As Cl is diatomic gas, no of molecules= 1.8808x10 23 x2=3.7616x10 23 molecules.
So, 3.14L of Ar at given condition contains more molecules.
If it is 11.07L of Cl 2 not 11.07gm and the condition is same as of Ar:
then obviously the higher volume of gas contains more molecules.
If it is 11.07L of Cl 2 at STP:
Then no of atoms for Cl 2 = (6.023x10 23 x11.07)/22.4=2.9765x10 23 atoms
= (2.9765x10 23 ?x2)=5.953x10 23 molecules.
So the Ar contains more no of molecules.
.
which sample contains more molecules- 3-14L of At at 85-0 degrees Cels.docx
1. which sample contains more molecules, 3.14L of At at 85.0 degrees Celsius and 1111mm Hg
pressure or 11.07 of Cl2?
Solution
I think it is 'Ar' not 'At'. But what is 11.07 of Cl 2 ? There is no unit.
If it is 11.07gm then:
Let, 3.14L of Ar at 85degree(358K) Cand 1111mm Hg pressure is equvalent to X L of Ar at
STP(298K and 76mm Hg).
So, we know (P 1 V 1 /T 1 )=(P 2 V 2 /T 2 )
Then, (1111*3.14)/358=(76* X )/298
or, X =38.209 L
According to Avogadros Law 22.4L gas at STP contains 6.023x10 23
no of atoms.
So, 38.209L of Ar contains (6.023x10 23
x38.209)/22.4= 1.02738x10 24
no of atoms.
For Ar no of atoms and no of molecules are same, as it is a innert gas.
Now atomic wt. of Cl 2 is 35.45
So, 35.45gm of Cl 2 contains 6.023x10 23
atoms
So for 11.07 gm of Cl 2 no of atoms= (6.023x10 23
x11.07)/35.45=1.8808x10 23
atoms.
As Cl is diatomic gas, no of molecules= 1.8808x10 23
x2=3.7616x10 23
molecules.
So, 3.14L of Ar at given condition contains more molecules.
If it is 11.07L of Cl 2 not 11.07gm and the condition is same as of Ar:
2. then obviously the higher volume of gas contains more molecules.
If it is 11.07L of Cl 2 at STP:
Then no of atoms for Cl 2 = (6.023x10 23
x11.07)/22.4=2.9765x10 23
atoms
= (2.9765x10 23
?x2)=5.953x10 23
molecules.
So the Ar contains more no of molecules.
