Question

Concentrations of solute in solvent, whether in saturated solutions or not, can be expressed in many ways (molarity, molality, ppm, percent by mass, etc.).

Molarity = moles of solute / liter of solution

molality = moles of solute / kg of solvent

ppm = mg solute / liter of solution

ppb = microgram of solute / liter of solution

%-by-mass = (grams of solute / grams of solution) x 100%

To convert from %-by-mass to Molarity, assume 100g of solution, use the solution density (in g/mL) to covert 100g solution to mL and then to L of solution, then divide that number up into the moles of solute to get Molarity.

To convert from Molarity to molality, assume 1 liter of solution and use the solution density (in g/mL) to convert 1 liter of solution to grams of solution, from which you subtract the grams of solute to get the grams of solvent. Convert grams of solvent to kilograms of solvent and divide up into moles of solute to get the molality.

To convert from molality to Molarity, assume 1 kg of solvent and add the grams of solute to the 1 kg of solvent to get the grams of solution, then use the solution density (in g/mL) to convert grams of solution to mL of solution and then convert that to liters of solution, which you then divide up into the moles of solute to get Molarity.

To convert from ppm to Molarity, remember that ppm = mg solute per liter of solution, so assume a liter of solution and use the solution density to convert to L of solution, then divide that up into the moles of solute to get Molarity.

Let’s start with a basic dilution problem first, which deals also with molar ratios (this is a refresher from chapter 4):

If 100.0 mL of 5.25 M Na3PO4 is diluted to a total volume of 250 mL in a volumetric flask, the sodium ion concentration, [Na+], in the resulting solution will be closest to…

A. 0.210 M

B. 0.630 M

C. 6.30 M

D. 0.021 M

Answer 1

If 100.0 mL of 5.25M Na3PO4 is diluted to a total volume of 250 mL in a volumetric flask, the sodium ion concentration, Na+|, in the resulting solution will be closest to A. 0.210 M B. 0.630 M C. 6.30 M D. 0.021 M Initial volume, V,-100.0 Initial molarity, M, -5.25 M Final volume, V, 250 mL Final molarity, M2 Use dilution law: mL 5.25 M x 100.0 mL M, x 250 mL So M1 -2.10 M SoNa Po2.10 M 1 mole of Na^PO4 gives 3 moles of Na* ions. Sol Na+ 3x2.10-6.30 M So the answer is C