a) How many moles of Al 2 (SO 4 ) 3 are required to make 45 mL of a 0.080 M Al 2
ID: 1057623 • Letter: A
Question
a) How many moles of Al2(SO4)3 are required to make 45 mL of a 0.080 M Al2(SO4)3 solution?
(Hint, molarity is moles per liter. You know the molarity. You are given the number of mL's, which can be converted to liters. Just set it up so the units cancel and you get "moles".)
_____________ moles Al2(SO4)3
b) What is the molar mass of Al2(SO4)3, to the nearest gram?
_____________ [grams/mole]
c) What mass of Al2(SO4)3 is required to make 45 mL of a 0.080 M Al2(SO4)3 solution?
(If you've got moles and molecular weight you can get to grams.)
______________ g Al2(SO4)3
d) How many moles of aluminum ions are present in the solution?
(This is easy. You know how many moles of Al2(SO4)3 there are from part a. Just think about how many aluminum ions there are per Al2(SO4)3.)
______________ mol
Explanation / Answer
Ans. a. Molarity = number of moles of solutes per L of solution = moles / L
Also, number of moles in a solution = Molarity x volume (in L) of the solution
Or, (moles / L) x L = moles
Thus, multiplying the molarity of a solution to the volume of the solution in Liter gibes the number of moles of the specified solute.
So,
The number of moles of Al2(SO4)3 required = Molarity x Volume (in L)
= 0.080 M x 0.045 L [1 L = 1000 mL]
= 0.0036 moles
Ans. b. Molar mass of Al2(SO4)3 = 342.13 gram/mol = 342 gram/mol (approx.)
Ans. c. Mass of Al2(SO4)3 required = moles of Al2(SO4)3 required x molecular mass of Al2(SO4)3
= 0.0036 moles x 342 g/mol = 1.2312 gram
And. d. Al2(SO4)3
1 mol of Al2(SO4)3 consists of 2 moles of Al ions (Al2 = 2 Al).
Therefore, the number of moles of Al in a sample is equal to twice the number of moles of Al2(SO4)3.
So, number of moles of Al = 2 x number of moles of Al2(SO4)3
= 2 x 0.0036 moles = 0.0072 moles
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