gen. chemistry question

[Erhatstil]

What is the molality of 85% phosphoric acid, H3PO4, if the density of the solution is 1.70 g/ml?

So if we have 85g H3PO4 per 15g H20 for the solution and need to get to the molal (moles/kg solvent), how do I relate it to the density?? Im a bit fuzzy on how to do this.

this is an achiever question and the answer is:
(1000)(1/1.70)(100/15)(1.70)(85/100)(1/98)m

thanks to anyone willing to help

---

Members don't see this ad.

 

[Erhatstil]

okay i think the density is irrelevant, which i should have noticed because it cancels out.

so (85g H3PO4/ 15g H2O)(1000g H20/ 1Kg)(1 mole H3PO4/ 98g H3PO4)=57.8m

anyone confirm this??? thanks a ton

 

[TKDentist]

looks good. you would have needed the density if they were asking for something with volume, such as molarity.

 

[UndergradGuy7]

You can also do this I think...

If it asks What is the molality of 85% phosphoric acid, H3PO4, if the density of the solution is 1.70 g/ml?

density is 1.70 g of solution / ml of solution

so if the solution is 85% h3po4 then 0.85*1.70=1.445grams of phosphoric acid in 1 ml of solution
If the solution is 85% phosphoric acid then 100-85=15% of solvent.
So 1.70*0.15=0.255 grams of solvent in 1 ml of solution = 2.55 x 10^-4 kg of solvent

So 1.445 grams of phosphoric acid/(98 g/mol) = 0.014744898 moles of phosphoric acid

so 0.014744898 moles phosphoric acid/2.44 x 10^-4 kg solvent = 57.823 molality

 

[Erhatstil]

that seems to work as well. i wouldn't have thought of that approach, good stuff!