Answer:
The required volume is 1.6 x 10³mL.
Explanation:
When we want to prepare a dilute solution from a concentrated one, we can use the dilution rule to find out the required volume to dilute. This rule states:
C₁ . V₁ = C₂ . V₂
where,
C₁ and V₁ are the concentration and volume of the concentrated solution
C₂ and V₂ are the concentration and volume of the dilute solution
In this case, we want to find out V₁:
C₁ . V₁ = C₂ . V₂
Answer:
a) 35.485 moles of Al(SO)
b) 35.485 moles of S atoms
c) Al atoms
d) 567.723 g of O
Explanation:
Let's define the following terms :
1 mol = elemental units
For example :
1 mol of oxygen atoms = oxygen atoms
Now, our compound has the following formula
Al(SO)
Where Al is aluminium
S is sulfur
And O is oxygen
All the subscripts are 1 so we can say the following :
1 molecule of Al(SO) has 1 atom of Al , 1 atom of S and 1 atom of O
In terms of moles :
1 mol of Al(SO) has 1 mol of Al , 1 mol of S and 1 mol of O
The molar masses of Al, S and O are
If we sum all the molar masses =
Finally, 75.046 g of Al(S0) is 1 mol of Al(SO) which contains 26.982 g of Al, 32.065 g of S and 15.999 g of O.
1 mol of Al(SO) contains 1 mol of Al, 1 mol of S and 1 mol of O.
Now we can calculate a),b),c) and d)
For a)
75.046 g of Al(SO) = 1 mol of Al(SO)
2663 g of Al(SO) = x
2.663 kg of Al(SO) contains 35.485 moles of Al(SO)
b) and c) 1 mol of Al(SO) molecules contains 1 mol of S atoms and 1 mol of Al atoms
We have 35.485 moles of Al(SO) molecules so
We have 35.485 moles of S atoms
And 35.485 moles of Al atoms
If 1 mol =
35.485 moles of Al have Al atoms
d) 75.046 g of Al(SO) contains 15.999 g of O
2663 g of Al(SO) contains x g of O
x = 567.723 g of O