[High School - Chemistry]Solutions

[Xception704]

I was sick for a day last week and they covered a whole chapter pretty much.

Here goes the question ... been trying to take a stab at it for half an hour ... thought I'd elicit help to see how to work it, then work out the rest using this one as a model.

How many g (grams) of solute would be required to make a 4.50m (for molality, not M for molarity) solution of H2SO4 in 1.00 kg H2O?

Let's see ... molality = moles of solute/kg of solvent

Any ideas?

Thanks in advance.

[dooey100]

Start by putting the numbers into the formula:
4.50m = moles/1.00kg

Multiply both sides by 1.00kg, and you get 4.50moles.

So you have 4.50 moles of H2SO4, and you need to turn this into grams.

moles * g/mole = g

you need the grams per mole of H2SO4, which you can find by adding up all the molar masses of 2 Hydrogen, 1 Sulfur, and 4 Oxygen

From there it's simple:

4.50moles * (the molar mass of H2SO4) = the mass of the H2SO4

Hope I helped.

If I didn't just tell me and I can go into more detail.

[Xception704]

Wow lol.

Yeah I went to school today and got it done.

I hate overthinking problems.

Thanks. =)

[Tasselfoot]

good tool to remember... EVERYTHING in basic chemistry boils down to stoichiometry.

[bender5]

Grr stoich is rather annoying I can show you a diagram that my chem teacher has us put on all of our papers to help us with conversions from Moles to Molarity to Liters/Kilograms .



Might be confusing at first, but it's pretty straight forward. If you have Molarity and Liters you multiply to get to Moles. If you have Moles and Molarity you divide to get to Liters. And if you have Moles and Liters you divide to get to Molarity. If you can just draw this chart out it makes things easier to just see the equation you need to do rather than memorize 3 different ones.

[dore]

Originally Posted by Tasselfoot

good tool to remember... EVERYTHING in basic chemistry boils down to stoichiometry.

why I made a 100 in Chemistry