[matilde79]

Hi there!
I have 2 problems to resolve:

1.Calculate how many ml of 12.4M HCI we need to make 5.00l of 0.100M HCI.
I started like this: V=(m/M)/c = (5000ml/36,450)/12,4M=11.624..  

2.How many g H2SO4 r present in 20.5ml of 0.100M H2SO4?

Can somebody please help me?

Matilde  

[jdurg]

What do you need help with?  We'll gladly help you along, but we won't do your work for you.  

[matilde79]

Just wanted to know if it's right or not.. and how to proceed..  

[jdurg]

Ahhh.  Okay.  I missed the part where you were showing your work.  (Coffee hasn't kicked in yet and I'm used to seeing people posting their homework here and expecting us to do it for them.  Sorry about that.   )

For your first problem, it's asking how much of a 12.4M (I'm assuming you mean HCl) solution you need to make 5.00 liters of a 0.100 Molar solution.  As all of this is volume related, there is no need for any masses to be used.

The first thing to do is go with your known data.  You know that you need 5.00 Liters of a 0.100Moles/L solution.  So how many moles is that?

Once you've got that number of moles, it's a simple ratio.  If 1000 mL of your 12.4 Molar solution contains 12.4 moles, how many mL of your 12.4 molar solution contains the number of moles in 5.00L of a 0.100M solution?

For question two we will need to use mass.  So what's the molar mass of H2SO4?  Okay, keep that value in mind.  Now how many moles of H2SO4 exist in 20.5mL of a 0.100Moles/1000mL solution?  Now it's just simple math.    (Remember that a 1.0M solution has 1.0 Moles of the solute in 1 Liter, a.k.a. 1000 mL of solution).

[matilde79]

Ok, solution nr.1:
c₁ v₁= c₂ v₂  .  12.4M * v₁ = 0,100M * 5l= 0,040ml

Is this right jdurg?  

Matilde

[jdurg]

Ok, solution nr.1:
c₁ v₁= c₂ v₂  .  12.4M * v₁ = 0,100M * 5l= 0,040ml

Is this right jdurg?  

Matilde

Allllllllllllllllllllmost there!  You just need to ensure that your units are correct.  In the equation you have 12.4Moles/Liter *x-Liters = (0.100Moles/Liter)*5Liters.  If the unit analysis is done properly, you'll have your answer.  

[Che-Comp]

Ok. Let me help you with the second question:

How many g H2SO4 r present in 20.5ml of 0.100M H2SO4?

We can attack this problem using dimensional analysis. We can use the formula weight as a conversion factor.  Also, we can use the molarity as a conversion factor. Another conversion factor needed is the conversion from ml to li.

Our target is g H2SO4. Usually it is easy to get the g if you have the mols since you can use the formula weight as a conversion factor. Do you know how to get the fomula weight of H2SO4? What is the unit? Right! g/mol.

So, how do we get the mols then? We can use the given volume and the molarity. What is the unit of molarity? Right! mols/liter.  So, what do you need to get the mols? You need the volume in liters. What do you need to do to the given volume in ml? Correct! Convert it to liters? How? What conversiton factor do you use? Exactly! 1000ml/1liter.

So, let's outline our process:

Vol of solution in ml ---> (conversion factor) --> Vol of solution in liters ---> (conversion factor)  --> mols of H2SO4 ---> (conversion factor) --> g H2SO4.

Now, all you have to do is to use the appropriate conversion factor. Make sure that the required unit is on the top of your ratios.  Example: 20.5ml (1 liter/1000ml).

I hope that helps!

[dolphinsiu]

1.Calculate how many ml of 12.4M HCI we need to make 5.00L of 0.100M HCI.

2.How many g H2SO4 r present in 20.5ml of 0.100M H2SO4?

Suggested Answers

1. 5 x  0.1/12.4  =  0.0403 L = 40.3 mL

2. no. of mole of H2SO4 = 0.1 x 20.5 /1000 = 2.05 x 10 -3 mol
    mass of H2SO4 = 2.05 x 10 -3  x ( 2 + 32 + 64 ) = 0.2 g

[matilde79]

so...

1)C₁ * V₁ = C₂ * V₂  -->  12,4M * V₁ = 0,100M * 5l= 0,040l --> 40ml

2)c=n/v  --> c=m/M*V  -->  0,100M= m/98,086*0,0205  --> m=98,086*0,0205*0,100M=0,2g

Thanx a lot!    Have a nice evening!

Matilde