Answer:
1. The starting assumption is an amount of 100 grams of solution.
2. It simplifies the problem, because the percentage tells directly the mass of solute present in 100 grams of solution and from that point the calculations are easy to do.
Explanation:
<em>Percentages are always based on the amount of total substance</em>, total solution if it is the case.
If it is mass percentage the amount of total substance is mass.
For instance, a 30% mass solution of sodium chloride means 30 grams of sodium chloride (solute) in 100 grams of solution (includes the solvent plus the solute).
- mass percent is mass % = (mass solute / mass solution) × 100%,
<em>Molarity</em> is number of moles of solute per volume in liters of solution:
- Molarity = (moles solute / liters solution)
In this case,<em> to convert from percentage to molarity</em> you start assuming a base of 100 grams of solution. For the example of the 30% sodium chloride solution, by assuming 100 grams of solution, you had 30 grams of solute, which you can convert into moles, and, by using the density of the solution (which you need to know), you convert the 100 grams of solution into volume (in liters). Then, you can divide the number of moles of solute into the number of liters of solution, having started with the assumption of 100 grams of solution.
<em>Molality</em> is the number of moles of solute per kilograms of solvent.
- Molality = moles solute / kg of solvent.
Again, <em>to convert from percentage to molality</em>, your first step is to assume a base of 100 grams of solution. Then, for the same example of 30% sodium chloride, convert 30 grams of solute to moles, and the 100 grams of solution into 70 grams of solvent (100g solution - 30 grams solute = 70 grams solvent), which is 0.070 kg. Finally, divide the moles of solute by 0.070 kg of solvent.
Then, assuming the amount of solution equal to 100 grams simplifies the problem because the percentage tells directly the amount of solute, which you can convert into moles making calculations easier than with other amount of solution.
