Question

How much ethanol (d = 0.789 g/mL), in liters, must be dissolved in water to produce 200.0 L of a 1.85 M solution?

Convert the concentration of ethanol in question 3A to ppm. Assume the density of the final solution is 1.00 g / mL. Describe why, or why not, parts per million is a useful unit to use for this particular system.

ACS grade 90. wt% Nitric Acid (d = 1.48g/mL) contains no more than 0.2 ppm Fe. What is the mass of iron in a 2.50 L bottle of nitric acid? If for an experiment you dilute the stock nitric acid above so that the concentration of iron is 4.0 ppb, what is the ratio of acid to total final volume needed?

Answer 1

Let us strat with the first part of the question :

Mol wt of Ethanol (C2H5OH) = 46

Moles of ethanol required = 200 L * 1.85M = 370.00 g needed
Now we are calculating the Volume
Vol = mass/density = 370.00 g / 0.789 g/mL = 468.94 mL or 0.468 L needed.

Generally we do not preffer ppm unit for this case.

Parts per million: The number of milligrams of solute per kg of solution = one ppm,

since 1 mg = 10-3 g and 1 kg = 103 g.

Assuming the density of water is 1.00 g/mL, 1 liter of solution = 1 kg and hence, 1 mg/L = 1 ppm. This is generally true for freshwater and other dilute aqueous solutions.

Parts per million concentrations are essentially mass ratios (solute to solution) x a million (106 ).

In this sense, they are similar to wt %, which could be thought of as parts per hundred (although nobody uses this term). Other variations on this theme include:

ppt – parts per thousand (used for common ions in sea water)

ppb – parts per billion (used for heavy metals and organics)

ppt – parts per trillion (used for trace metals and trace organics)