Gchem q

[Ruthless]

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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? (FW of H2SO4 = 98
g/mol)
A. 1.6
B. 3.0
C. 4.0
D. 5.0
E. 6.25

The answer is c. but i can not seem to grasp how to solve this using specific gravity? Can any explain this well and maybe post other specific gravity questions? I am sure many others are having problems with this. Thanks

 

[Ruthless]

i guess no one understands it either.

 

[dufran3]

Yeh sorry man, GChem isn't my thing. I'm not sure how to do that problem either. Someone on here knows how though. I'm interested to find out how to solve the problem as well!

 

[UCfan]

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

The answer is c. but i can not seem to grasp how to solve this using specific gravity? Can any explain this well and maybe post other specific gravity questions? I am sure many others are having problems with this. Thanks

Specific gravity problems?I do not know what you mean.From specific gravity you can find density (That's all I remember).

The statement says
you have
39.2 g H2SO4/100 g solution
you need to know that density of water is 1 g/cm^3. So density of the solution is 1.25 g/cm^3.

(39.2 g H2SO4/100 g solution)*(1.25 g solution/1 cm^3)*(1000cm^3/1L)=
490 g H2S04 /L

From here(you need to use molecular weight)
490 g H2SO4/L *(1 mol/98 g H2SO4)=5 mol/L
You need (0.20M*0.1L=0.02 mol H2SO4)
So
0.02/5=0.004L= 4ml
Hopefully this helps!

 

[dufran3]

It's hard to keep everything straight during those calculations. Easy to get lost in what exactly you are finding.

 

[UCfan]

It's hard to keep everything straight during those calculations. Easy to get lost in what exactly you are finding.

Which part do you have difficulties to understand?

 

[dufran3]

No no, I followed the calculations fine. I even wrote them down and did the calculation myself. I'm just commenting that it is a problem requiring really 3 separate steps, and w/out your helpful step by step calculations it "could" be easy to get lost in it.

The problems is challenging for me because you have to immediately remember that the 39.2% is just 39.2 g/ 100 g. Which seems obvious but I always forget that part. Also, knowing the units for the specific gravity is tricky.

 

[Ruthless]

Specific gravity problems?I do not know what you mean.From specific gravity you can find density (That's all I remember).

The statement says
you have
39.2 g H2SO4/100 g solution
you need to know that density of water is 1 g/cm^3. So density of the solution is 1.25 g/cm^3.

(39.2 g H2SO4/100 g solution)*(1.25 g solution/1 cm^3)*(1000cm^3/1L)=
490 g H2S04 /L

From here(you need to use molecular weight)
490 g H2SO4/L *(1 mol/100 g H2SO4)=4.9 mol/L
You need (0.20M*0.1L=0.02 mol H2SO4)
So
0.02/4.9=0.00408L= 4ml
Hopefully this helps!

Thanks man! perfect.