i cant get this question!

[jp370]

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1. it is off the kaplan midterm question #58

An aqueous sulfuric acid solution is 39.2% H2SO4 by
mass and has a specific gravity of 1.25. How many
milliliters of this solution are required to make 100 mL
of a 0.20 M sulfuric acid solution? (FW of H2SO4 = 98
g/mol)
A. 1.6
B. 3.0
C. 4.0
D. 5.0
E. 6.25
plz help i only have three weeks to dat and feel so dumb!

 

[doc toothache]

http://forums.studentdoctor.net/showpost.php?p=5952159&postcount=3

 

[doc3232]

Whats specific gravity and is it always greater than 1?
Can you please explain it in context with the problem too?
THANKS

 

[ChemE87]

Specific gravity is the density of matter compared to the density of water, which is 1.00 at 4degC and 1atm. Specific gravity also doesn't have any units, since it's a ratio, and can be used in any measurement system. All you have to know is the density of water and multiply by the specific gravity of the material/liquid you are using to get a number with units (ex. SG=1.25 so it's density would be 1.25g/ml, 1.25kg/L, 78.0lb/ft^3, etc).

And no, specific gravity is not always greater than 1. Things wouldn't float on water if that was true.

 

[TeamGuo]

ok, what I did was

98 g/mol divide 1.25 g/ml = ml/mol
then flip it to get mol/ml , multiply 1000 to get mol/L

it's just an M1V1 = M2V2 question

once you get M1 from above, multiply .392 to get the final M1
plug in solve for V1

answer is 4.0

 

[dentalplan]

1. it is off the kaplan midterm question #58

An aqueous sulfuric acid solution is 39.2% H2SO4 by
mass and has a specific gravity of 1.25. How many
milliliters of this solution are required to make 100 mL
of a 0.20 M sulfuric acid solution? (FW of H2SO4 = 98
g/mol)
A. 1.6
B. 3.0
C. 4.0
D. 5.0
E. 6.25
plz help i only have three weeks to dat and feel so dumb!

Does you understand how Doc toothache did it though?

The general concept is....you have an original solution, and you want to make a new solution by doing a few things.........

(2 steps)

1.) you must first find the number of grams of H2SO4 in a 0.2 M, 100 mL solution of H2SO4 (the new solution). So do 0.2 M x 0.1 L x 98 grams. You get 1.96 grams. On the real DAT, since you don't have a calculator you have to compute this quick, and since you may not be sure it is 1.96, just round to 2 grams for quickness.

2.) next you need to take your original solution (which is ~39 % H2SO4 by mass), and find out how many milliliters of it you would need, in order for there to be 1.96 grams of H2SO4.

To actually get 1.96 grams from the original solution, you have to start off somewhere.😀

So to start off, you know that the original solution has 1.25 grams of substance present in one milliliter. This is NOT the amount of H2SO4 present though. The amount of H2SO4 present is actually 39% of the 1.25 grams.



---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.25 grams x (39 grams H2SO4/100 grams of solution) = 0.49 g H2SO4. 39% of 1.25 grams is 0.49 grams. 0.49 grams is the amount of H2SO4 present in one milliliter of the original solution...and thus since we need 1.96 grams of H2SO4 in our new solution, we multiply 0.49 by four. 0.49 grams per milliliter x 4 milliliters= 1.96 grams. If you notice, we just need 4 milliliters of the original solution, in order to get that 1.96 g value. Now we are done, since we figured out how many milliliters of the original solution we needed in order for us to actually get 1.96 grams of H2SO4.------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Hope that cleared it up for you.