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Molarity, Volume and Moles

Submitted by samanthadoyle on Wed, 10/24/2007 - 01:55

Calculate each of the following quantities:

(a) volume of 14.7 M sulfuric acid that must be added to water to prepare 2.00 L of a 0.271 M solution

(b)  molarity of the solution obtained by diluting 75.7 mL of 0.231 M ammonium chloride to 0.250 L

(c)  volume of water added to 0.150 L of 0.0252 M sodium hydroxide to obtain a 0.0100 M solution (assume the volumes are additive at these low concentrations)

(d)  mass of calcium nitrate in each milliliter of a solution prepared by diluting 56.5 mL of 0.726 M calcium nitrate to a final volume of 0.100 L

I am completely hopeless with these problems.  :-[  I keep trying different things, and nothing seems to come out in the right units.  Please help me! 


Calculate each of the following quantities:

(a) volume of 14.7 M sulfuric acid that must be added to water to prepare 2.00 L of a 0.271 M solution

step1. you have to find the number of moles needed to make the solution.

recall: in word, we can state 0.271 M as there is0.271 moles of acid per 1 liter of solution... in figure...

0.271 moles / 1 liter so we have

# of moles needed = (0.271 moles / 1liter) x (2 liters) = solve for that (1)

step 2. find the volume needed to get your computed # of moles (1) from the 14.7 M sulfuric acid

that is given by

volume needed = (1 liter / 14.7 moles) x ( # of moles you computed)

(b)  molarity of the solution obtained by diluting 75.7 mL of 0.231 M ammonium chloride to 0.250 L

recall that

1 M = (1 mole / 1 liter) = (1 mmol / mL) {millimoles}

so your porblem will be like this

(75. 7 ml) x (0.231 mmol / mL)
250 mL {since .250 L is 250 mL)

(c)  volume of water added to 0.150 L of 0.0252 M sodium hydroxide to obtain a 0.0100 M solution (assume the volumes are additive at these low concentrations)

ok, this one requires your analysis...

recall that:

0.0100M soln is = 0.0100 moles / L

if you have 0.150 L of 0.0252 M of NaOH

it means you have

0.0252 mole / 0.150 L = j

compute that so that you will have the units

j mole / liter

logic: if i have j moles per liter and i want it to be 0.0100 moles per liter, what should be the volume of water should i add?

here it goes... (algebra)

M = mole / L
and
0.0100 M = 0.0100 mole / L

so

0.0100 mole / liter = 0.0252 moles / x liter

rearranging we have

x liter = 0.0252 moles / (0.0100 mole/liter)

compute that and you will have the total volume of the solution,

but you already have an initial volume 0.150 L, so subtract it form X liter... and you will have the needed volume of water

(d)  mass of calcium nitrate in each milliliter of a solution prepared by diluting 56.5 mL of 0.726 M calcium nitrate to a final volume of 0.100 L

again:

get the final Molar concentration M of the solution after the dilution

M = mole / L or mmol / mL

so get the number of mmol per mL

multiply it to the molar mass of calcium nitrate (Ca = 40, N = 14.01 and O = 16)

go and solve... you can do that! good luck!


You said that you keep trying different things, but you can't get the right units. Are you getting the right answer, just with the decimal places off because of the units? If thats the case, you just have to recognise how many, say mL are in a L, and do the proper conversion. How have you been approaching the problems? We might be able to get a better idea of where you are going wrong if you could just show us how you are solving for example A. Once, we get that figured out, we can all pitch in and lend a helping hand to get you to better understand it.


samanthadoyle wrote:

Calculate each of the following quantities:

(a) volume of 14.7 M sulfuric acid that must be added to water to prepare 2.00 L of a 0.271 M solution

(b)  molarity of the solution obtained by diluting 75.7 mL of 0.231 M ammonium chloride to 0.250 L

(c)  volume of water added to 0.150 L of 0.0252 M sodium hydroxide to obtain a 0.0100 M solution (assume the volumes are additive at these low concentrations)

(d)  mass of calcium nitrate in each milliliter of a solution prepared by diluting 56.5 mL of 0.726 M calcium nitrate to a final volume of 0.100 L

I am completely hopeless with these problems.  :-[  I keep trying different things, and nothing seems to come out in the right units.  Please help me! 

So...here is how we solve.

(A) It's best to start off with the most information given. In this case, it is 2.00L and 0.271M. Now, we can use stoichemetry to solve the problem. Keep in mind, that we first want to find moles, but ultimately, we want to find volume (in Liters) for our final answer.

2.00L (0.271 mol H2SO4)  = 0.542 mol H2SO4
                (1L)

Now that we have found moles, we can now use our 0.542 mols with our lesser known information of 14.7M to find volume in Liters. This needs to be arranged so that mols will cancel out leaving us with Liters.

0.542 mol    (1L)  = 0.0369 L H2SO4
              (14.7 mol)

(B) We are required to find the molarity (M). Again, start with the most known information available. In this case, it is 75.7 mL and 0.231M. Before you start, take a look at your units, particularly 75.7 mL. You need to convert this to L.

Converting 75.7 mL to L
75.7 mL  (1 L)      = 0.0757 L
          (10-3 mL)

Now we can start by solving for moles.
0.0757 L  (0.231 mol NH4Cl) = 0.0175 mol NH4Cl
                (1 L)

Remember that Molarity (M) is simply M = moles solute
                                                volume of solution in Liters

With that being the case, we have found moles (0.0175 mol), and we have the volume of our lesser of information (0.250 L). Now, we just have to use the equation for Molarity, and we will come out with our answer in M.

M = (0.0175 mol) = 0.0700 M NH4Cl
      (0.250 L )

(C) This problem is fairly identical to problem (A), being that it approaches the situation in the same way. Again, we are required to find volume (Liters). Start with the most known information. In this case it is 0.150 L and 0.0252 M

0.150 L  (0.0252 mol NaOH) = 0.00378 mol NaOH
              ( 1 L )

Now that we have moles, use that answer with the lesser of the known information, to find our volume in Liters.

0.00378 mol  ( 1 L )  = 0.0378 L NaOH               
              (0.100 mol)

(D) Ok, like most chemistry books, and professors, they generally give you the easy ones first, and then slam you with one that requires more work then you just got done doing. This is an example of one compared to problems (A-C). So, lets start. There are a few things you are going to need to know in order to work this problem out. First, it asks for the answer to be in the form of mass. Mass is always in grams (g) unless specified otherwise. Next, you need to know how to write calcium nitrate in its elemental form. Finally, the rest is simply the same process as before.

Lets start with writing out calcium nitrate in its elemental form, because this will be helpful when finding the molar mass, which is required at the end of the problem.
Ca2+ + NO3- ---> Ca(NO3)2 

Now that we have calcium nitrate written out, we can start our conversions. Like the other problems, start with the most information known. In this case it is, 56.5 mL and 0.726 M (Make sure to change mL to L)
56.5 mL  ( 1 L )      = 0.0565 L
            (10-3 mL)

Next,
0.0565 L  (0.726 mol Ca(NO3)2) = 0.0410 mol Ca(NO3)2
                ( 1 L )

Remember that I mentioned we would have to find the molar mass of calcium nitrate? Well, this is where we need to find the molar mass in order to covert from mol --> grams. The molar mass of each element is found on the periodic table. The molar mass of calcium nitrate is,

1 Ca    1(40.08) = 40.08 g/mol

2 N    2(14.007) = 28.014 g/mol

6 O    6(15.999) = 95.994 g/mol

Add the weights together, and the forumla weight for calcium nitrate is = 164.008 g/mol

Now, proceding with the stoichiometry.

0.0410 mol Ca(NO3)2  ( 164.008 g Ca(NO3)2) = 6.72 g Ca(NO3)2
                                ( 1 mol Ca(NO[sub]3)2)

I hope that I was able to also help you out in solving for the problems. Please do not just copy and use in your notebook to turn in, rather use my answers and set-ups to guide you along the way, as this well help you far more than just using my answers for yours strickly to call the assignment done. If you have any questions, or I got something wrong that the answer key suggests is right, please let me know. Hope this helps. Best of luck!! 8)