Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: DilutedBrain on February 13, 2020, 08:58:22 AM

The Problem:Calculate the number of grams of solute necessary to prepare 700 grams of 0.6m H_{2}SO_{4} solution.
The attempt: I tried to find the answer using the formula:
grams solute = molality * Molecular weight of H_{2}SO_{4}*L solvent
...until I realized there is no solvent in the problem. I tried to find the kilograms of solvent but there's no way I can find it without grams of solute or the weight percent. What formula should I use? Should I convert the molal concentration into molar concentration then find the grams of solute from there?

Calculate the mass of sulfuric acid needed to make 0.6m from 1kg of water, calculate the mass of this solution, calculate the scaling factor to 700g and apply it to the previously calculated mass of sulfuric acid.
Show your result.

So... this is what I did
1. Calculate the mass of Sulfuric acid...
mass H_{2}SO_{4} = O.6m * 98 g/mol (MW of H_{2}SO_{4}) * 1000 grams of water
I got 58,800 g H_{2}SO_{4}
2. Calculate the mass of this solution
I assumed the H_{2}SO_{4} solution so I based the answer to the given of 700g H_{2}SO_{4} solution. I don't get this one so I assumed this completely.
3. What I did for the scaling factor is this
58,800 grams/700 grams = 84 grams
...did I got this one right? or I failed? ???

The scaling (down) factor is 700/1058.8

Ohh... May I ask where the 1058.8 came from?

1000+58.8

1000+58.8
ohh okay thank you.
So I got the answer of 0.66. Then I multiplied it to the mass of the H_{2}SO_{4} and got 38.8 kg. Is this it or there's another step that I should take? ???

How you diminish 58.8 g to 38.8 kg

How you diminish 58.8 g to 38.8 kg
58.8 * 0.66 = 38.8

The rounding is a bit incorrect.

The rounding is a bit incorrect.
Ohh... I got 38.808. I'll put it in 2 sig fig just to be sure.
So... I got 39.

0.6*H2SO4*700/(1000+0.6*H2SO4) = 38.9029
so 40, 39 or 38.9 depending on the number of significant digits (1, 2 or 3)

Ohh okay I get it now. Thank you so much for the help :)

Check this out: https://www.facebook.com/ChemBuddyCC/videos/627397594716950/ :)

you know this
molality = 0.6m H2SO4 = moles solute / kg solvent
mass solution = 700g
mass solvent = mass solution  mass solute
you could always put in the effort and derive an equation to relate m, mass solution
mass solvent, etc.... but... the easy way is just to assume a known value of something
then calculate away. Since this is molality, let's assume we have 1 kg of solvent
in which case, we would have
1kg solvent
1kg solvent * 0.60 mol H2SO4 / kg solvent = 0.60 mol H2SO4
mass H2SO4 mol = 0.60 * 98.09 g/mol = 58.85g H2SO4
total mass solution = 1kg + 0.05885kg = 1058.85g
and now we know 58.85g H2SO5 = 1058.85g solution... the rest is trivial via dimensional analysis
700g solution 58.85g H2SO4
 x  = 38.90g H2SO4
1 1058.5g solution
*********
remember that trick of assuming 1kg of solvent or 1L of solution, etc. It works very well for molarity / molality type problems.

and now we know 58.85g H2SO5 = 1058.85g solution... the rest is trivial via dimensional analysis
700g solution 58.85g H2SO4
 x  = 38.90g H2SO4
1 1058.5g solution
minor printing errors

oops... my apologies
700g solution 58.85g H2SO4
 x  = 38.90g H2SO4
1 1058.85g solution
and it's H2SO4. sulfuric acid not persulfuric acid
good catches.