© R. Paselk 2008
1. What is the molarity of a solution made up by dissolving 46.7 g of Calcium Nitrate (FW= 164.1) in enough water to make up a volume of 500.0 mL?
First note that we don't even need the formula of calcium nitrate since the FW is given! Can just go ahead and calulate the number of moles:
Recall that FW has units of g/mol, therefore want to divide the mass by the FW to get units of mol.
moles = (46.7 g)/(164.1 g/mol) = 0.28458 mol calcium nitrate
The molarity (M) is defined as mol/L, and 500.0 mL = 0.5000 L so:
(0.28458 mol)/(0.5000 L) = 0.56917 mol/L = 0.569 M
2. What is the molarity (M) for chloride ion in a solution made up by dissolving 46.7 g of Aluminum Chloride (FW = 133.33) in enough water to make up a volume of 500.0 mL.
First notice that the question is NOT asking for the molarity of aluminum chloride, rather it is asking the molarity of the chloride ION.
So the first thing we need to do (so we don't forget later) is find the ratio of moles chloride ion per mol aluminum chloride, and for that we need the formula of aluminum chloride.
From the Periodic Table we see aluminum is an elemental ion in Group III while chloride is an elemental ion in Group VII:
IA IIA IIIA IVA VA VIA VIIA VIIIA IIIB IVB VB VI VIIB VIIIB IB IIB Al Cl Therefore Al has a charge of 3+ (Al+3) while chloride has a charge of 7-8 = -1 so chloride = Cl –.
Since the charge on any compound MUST be zero there are three chlolrides/aluminum, giving us a formula of AlCl3.
Now can go back and find the molarity of the aluminum chloride:
(46.7 g)/(133.33 g/mol) = 0.35026 moles, and since 500.0 mL = 0.5000 L
(0.35026 mol)/(0.5000 L) = 0.7005 mol/L = 0.7005 M aluminum chloride
From the formula we can see that there are three chlorines for each formula, therefore there are three moles of chlorine for each mole of aluminum chloride, giving:
(0.7005 M Cl –)(3 Cl/AlCl3) = 2.101 M Cl – = 2.10 M Cl–
3. What is the molarity of ions in a solution is made up by dissolving 7.67 g of Calcium Iodide in enough water to make up a volume of 250.0 ml.
First note that the question asks for TOTAL ions, that is molarity of calcium ions plus mlarity of iodide ions.
From the formula found from the charges of the ions from the Periodic Table, CaI2, there are three ions/Formula.
Next we need the FW to find the moles of calcium iodide:
(40.08 + 2x 126.9) g/mol = 293.88 g/mol
And molarity of solution is:
(7.67 g)/(293.88 g/mol) = 0.026099 mol, and since 250.0 mL = 0.2500 L
( 0.026099 mol)/(0.2500 L) = 0.1044 mol/L = 0.1044 M
Molarity of total ions is then:
(0.1044 M)(3 mol ions)/(mol CaI2) = 0.31319 M = 0.313 M
4) How much iron(III) sulfate must be weighed out to make up 200.0 mL of a 25.0 mM solution?
Here we have to find two things, the FW of the compound and the number of moles needed. Let's start with the moles:
Moles iron(III) sulfate needed is:
Recall 1L = 1000mL and mM = 10–3 M = 10–3 mol/L
(25.0 x 10–3 mol/L) (200.0mL)(1L/1000mL) = 5.000 x 10–3 mol iron(III) sulfate
Now need FW, and thus formula. Recall iron(III) = Fe+3 & sulfate = SO4–2, so formula = Fe2(SO4)3
So FW:
2(55.84) + 3(32.06 + 4 x 16.00) g/mol = 399.86 g/mol
Finally multiply moles by FW:
( 5.000 x 10–3 mol)(399.86 g/mol) = 1.9993 g = 2.00 g
5. How much Zinc chloride is needed to make up 1.500 L of a 0.250 M solution?
(1.500 L)(0.250 mol/L)(65.41 + 2 x 35.45) g/mol = 51.116 g ZnCl2 = 51.1 g ZnCl2
6. What volume of a 0.9954 M solution of sodium phosphate is needed to make up 250.0 mL of a 15.0 mM solution?
The thing to keep in mind in dilution problems is that the number of moles before = the number of moles after, or
moles = moles
Molarity1 x Volume1 = Molarity2 x Volume2
M1V1 = M2V2
In our problem our initial molarity is 0.9954 ( M1= 0.9954 M) with an unknown volume (V1 = ?), and final molarity is 15.0 x 10–3 M (M2 = 15.0 x 10–3 M) while the volume is 250.0 mL (V2 = 250.0 mL). So,
M1V1 = M2V2
(0.9954 M)(V1) = (15.0 x 10–3 M)(250.0 mL)
rearranging, dividing both sides by M1
V1 = V2 x (M2)/(M1), substituting,
V1 = (250.0 mL) x (15.0 x 10–3 M)/(0.9954 M) = 3.767 mL = 3.77 mL
7. What volume of a 2.568 M solution of hydrochloric acid is needed to make up 250.00 mL of a 0.500 M solution?
As above, start with moles = moles,
M1V1 = M2V2, or
V1 = V2 x (M2)/(M1), substituting,
V1 = (250.00 mL) x (0.500 M) / (2.568 M) = 48.676 mL = 48.7 mL
© R A Paselk
Last modified 9 March 2015