Chemistry
Elements and Compounds

# How many grams of K2SO4 would be needed to prepare 4.00 L of a 0.0510 M solution?

The needed mass is 35,549 g.

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## Related Questions

Molarity = moles of solute/Liters of solution ( 100 mL = 0.1 Liters)Moles of solute (K2SO4) = Liters of solution * MolarityMoles K2SO4 = 0.1 Liters * 0.1 M= 0.01 moles K2SO4 (174.27 grams/1 mole K2SO4)= 1.7 grams potassium sulfate=======================Add that many grams potassium sulfate to your 100 mL.

K2SO4 is considered a strong electrolyte. This is because it is a good conductor that can totally or partially dissociate or ionize in a solution.

Barium chloride solution: Ba2+ and Cl-. Potassium sulfate: K+ and (SO4)2-.

There is 1 sulfur atom in potassium sulfate (K2SO4).

K2SO4 is the chemical formula for Potassium sulfate.

K2SO4 ==&gt; 2K^+ and SO4^2- ions. So, each mole of K2SO4 produces 3 moles of ions.2.40 moles K2SO4 x 3 moles ions/mole K2SO4 = 7.2 moles of ions.

Two potassium ions. 2K(+) and One polyatomic ion of sulfate SO4(2-)

2KOH + H2SO4 --&gt; K2SO4 + 2H2O You get potassium sulphate, a salt, and water.

Molar mass of K2SO4 = 174.2592 g/mol

K2SO4 is potassium sulfate, and it is soluble in water.

Potassium sulfate has a molar mass of 174.259 grams per mole. In15 grams there are then 86.1 millimoles of K2SO4.

BME: (NH4)2SO4 + KOH --&gt; 2 NH4OH (or NH3 + H2O) + K2SO4 Net Ionic: NH4+ + OH- --&gt; NH3 + H2O (or NH4OH). :)

A fair question and some tricky points. Firstly, K2SO4 can be seen as 2K+[SO4]. If you put it into water, you get 2 potassium ions for each K2SO4. That's important. The rest is just wrestling with the numbers. We have 200cm3 of 2.0M K2SO4. How many moles is that? Well we'd have 2 moles in 1000cm3, so... (1000/200 = 5) we have 5 times less volume meaning 5 times fewer moles (2.0/5 = 0.4) so we have 0.4 moles K2SO4. Good. Now we put that into 800cm3 of water. There's a trick, here! It's not 0.4 moles into 800cm3 water, because our 0.4moles are ALREADY in their own 200cm3 of water. So remember to add the volumes up, too. But this just makes things so much easier for us! Because we end up with 1 litre which is the volume component of the molarity unit. (M = g/dm3 or g/L, same thing) So it's 0.4 moles of K2SO4 in 1litre of water, meaning 0.4M K2SO4. Finally....we want K+ ions, not K2SO4. Remember we had two K+ ions for every K2SO4? 2 x 0.4 = 0.8. So our answer is: we have 0.8M K+ solution in the end.

10.0 moles K2SO4 (6.022 X 1023/1 mole K2SO4) = 6.02 X 1024 atoms of potassium sulfate ==========================

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