G-chem question

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riro2

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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?

Answer: 4 ml

Can someone please explain how to do this problem. Its a practice test question that i cannot seem to solve.

Thanks
 
Straight up dimensional analysis...nothing to it:

100 mL x (0.2 mol H2SO4 / 1000 mL solution) x (98 g H2SO4 / 1 mol H2SO4) x (100 g solution / 39.2 g H2SO4) x (1 mL stock/ 1.25 g stock) = 4 mL stock

In order: Desired Volume, Desired Molarity, Molar Mass of H2SO4, %Mass in Stock, Density of Stock
 
In these type of problems, do we always begin with the desired volume/molarity or whatever is desired and then go from there using the dimensional analysis?
 
In these type of problems, do we always begin with the desired volume/molarity or whatever is desired and then go from there using the dimensional analysis?

Yeah, that's the best way to do it. That way you know which orientation to write the following conversions in order to make everything cancel out correctly. If you started with the molar mass, for example, you wouldn't know whether to write g/mol or mol/g...it would still work out, but everything ends up inverted and its harder to keep track it.
 
stupid question, but I don't understand the last calculation density of stock?
 
stupid question, but I don't understand the last calculation density of stock?

Correct me if I am wrong, stock in this case means the solution. We are given the specific gravity of solution and are required to find the milliliters of solution.
 
stupid question, but I don't understand the last calculation density of stock?

Well I just use the terms "desired" and "stock" to distinguish the solution we're trying to make from our starting solution. Normally in a lab, you have chemical stock of the reagents that you frequently use, and anytime you need to make a batch at working concentration you take some concentrated stock and dilute down.

Also, in the problem statement, "specific gravity" is just another term for density in g/mL.
 
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