Asked by Sam
how would you prepare 250 ml of 0.69m al2(so4)3 solution from a 2.0 al2(so4)3 stock solution
Answers
Answered by
DrBob222
2.0 what? m
Is that 0.69m or 0.69M? I assume 0.69m
That means 0.69 mol/0.250 kg solvent.
0.69 x molar mass Al2(SO4)3 = approx 60g.
Then the solvent has a mass of 250 g for a total mass of approx 250+60 = approx 0310 g. What's the density of the solution? volume = mass/density = ? = approx 310 mL or 0.310 L.
Then use c1v1 = c2v2.
Is that 0.69m or 0.69M? I assume 0.69m
That means 0.69 mol/0.250 kg solvent.
0.69 x molar mass Al2(SO4)3 = approx 60g.
Then the solvent has a mass of 250 g for a total mass of approx 250+60 = approx 0310 g. What's the density of the solution? volume = mass/density = ? = approx 310 mL or 0.310 L.
Then use c1v1 = c2v2.
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