[travibe]

ok im working on this problem about molar concetration and it seems easy enough, i just need a nudge in the  right direction...i  know the equatio nfor molar concetration is Concentration = Amount of X/Volume of sol'n.  but i am given to amounts and not sure which to use....a 50 ml sample with a mass of 50.320g is evap. to dryness, the reulting residue is .453g. Find the molar conc. of the NaCl sol'n.  Do i use the evaporated amount of the mass ebfore?

Either way, i get 9.06x10^-3g/mL if i use the .453 and 1.0064g/mL if i use the 50.320g.  Not sure which way to take it.

[Borek]

Since when molar concentration is given in g/mL?

http://www.chembuddy.com/?left=concentration&right=molarity

[travibe]

my apologies, i now see that it is mol/l...

Do i ignore the evaporated information? (im assuming not) ... i dont know where to put it in the question to solve for concetration.

btw,
Molarity = Mol/v
Volume = Mol/molarity
Mol = Molarity x Volume

are these conversions right?

[travibe]

ok tell me if i am on the right track.

Molarity = Mols of Solute/Liter(s) sol'n

.453g of NaCl
Molar Mass of NaCl is 58.48g

.453g/58.48g/mol = .00775 mol = 7.75x10^-3mols

Molarity of NaCl = (7.75x10^-3 mols NaCl) / .050 L

M = .155 mol/L

Is this anywhere near right?

[Borek]

OK

Although original data from the question seems to be wrong - mass of the solution should be not 50.320, but 50.230 g. Perhaps a typo?

Checked using CASC in case anybody wonders