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### Topic: Calculating number of grams of solute from molal concentration  (Read 1181 times)

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#### DilutedBrain

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• Mole Snacks: +0/-0 ##### Calculating number of grams of solute from molal concentration
« 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 H2SO4 solution.

The attempt: I tried to find the answer using the formula:

grams solute = molality * Molecular weight of H2SO4*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?

#### AWK

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« Reply #1 on: February 13, 2020, 09:14:47 AM »
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.
AWK

#### DilutedBrain

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« Reply #2 on: February 13, 2020, 10:07:36 AM »
So... this is what I did

1. Calculate the mass of Sulfuric acid...

mass H2SO4 = O.6m * 98 g/mol (MW of H2SO4) * 1000 grams of water

I got 58,800 g H2SO4

2. Calculate the mass of this solution

I assumed the H2SO4 solution so I based the answer to the given of 700g H2SO4 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? #### AWK

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« Reply #3 on: February 13, 2020, 10:13:54 AM »
The scaling (down) factor is 700/1058.8
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#### DilutedBrain

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« Reply #4 on: February 13, 2020, 10:17:46 AM »
Ohh... May I ask where the 1058.8 came from?

#### AWK

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« Reply #5 on: February 13, 2020, 10:19:58 AM »
1000+58.8
AWK

#### DilutedBrain

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« Reply #6 on: February 13, 2020, 10:27:15 AM »
1000+58.8
ohh okay thank you.

So I got the answer of 0.66. Then I multiplied it to the mass of the H2SO4 and got 38.8 kg. Is this it or there's another step that I should take? #### AWK

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« Reply #7 on: February 13, 2020, 10:29:21 AM »
How you diminish 58.8 g to 38.8 kg
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#### DilutedBrain

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« Reply #8 on: February 13, 2020, 10:35:26 AM »
How you diminish 58.8 g to 38.8 kg

58.8 * 0.66 = 38.8

#### AWK

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« Reply #9 on: February 13, 2020, 10:36:02 AM »
The rounding is a bit incorrect.
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#### DilutedBrain

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« Reply #10 on: February 13, 2020, 10:41:18 AM »
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.

#### AWK

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« Reply #11 on: February 13, 2020, 10:44:10 AM »
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)
AWK

#### DilutedBrain

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« Reply #12 on: February 13, 2020, 10:46:11 AM »
Ohh okay I get it now. Thank you so much for the help #### Borek ##### Re: Calculating number of grams of solute from molal concentration
« Reply #13 on: February 13, 2020, 05:50:52 PM »
Check this out: https://www.facebook.com/ChemBuddyCC/videos/627397594716950/ ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### MNIO

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• Mole Snacks: +12/-3 ##### Re: Calculating number of grams of solute from molal concentration
« Reply #14 on: February 14, 2020, 03:10:21 PM »
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.