Chemistry II concentration problem....?

Question

A solution is prepared by dissolcing 23.7 g of CaCl2 in 375g of water. The density of the resulting solution is 1.05 g/mL. the concentration of the Cl ion in this solution is ______ M?

Answer 1

Concentration has units called Molar (M) or Moles/Liter. So to answer the question we need moles of Cl per Liter of solution. Moles of Cl can be found by converting mass of Cl to moles using the molecular weight. 23.7 g of CaCl2 is (23.7g / 111.03 g/mol) 0.21345 mol CaCl2. here 111.03 is the molecular weight of CaCl2. Now the molecular formula, CaCl2, tells us that there are two moles of Cl in one mole of CaCl2. So we must multiply the moles of CaCl2 by two to get the moles of Cl. 2 x 0.21345 = 0.4269 mol Cl. Now all we need is the volume of solution. We know that the final density is 1.05 g/mL. This is a slight increase from pure water (1.00 g/mL) due to the addition of the CaCl2. All together we have 375+23.7 = 398.7 total grams of solution. Dividing the mass by the density will give volume (recall that density = mass/volume). So 398.7 g / 1.05 g/mL = 379.7 mL. This final volume is a bit higher than the original after adding the CaCl2. Now we have both moles and volume so we can calculate the concentration. Finally 0.4269 mol / 379.7 mL = 0.001124 mol/mL. Converting to mol/L is 1.124 M.


Note: sometimes the volume change can be neglected. This is the case for small concentrations. In this example the volume change was small but still relevant to get the point across. Interestingly, in some cases, the volume can even decrease after dissolving a substance (such as urea) in water.

Answer 2

Assuming the density of pure distilled water is 1 g/mL. The % by mass of CaCl2 is therefore 5% - 1 g of water and 0.05 g of dissolved CaCl2.

In 1 L = 1 dm^3 of calcium chloride solution, the total mass is 1050 g of which 50 g is dissolved CaCl2 solid.

CaCl2 has a Mr = 40 + 2(35.5) = 111
Hence 111 g = 1 mole of CaCl2

50 g = [50/111] mole.
CaCl2 = Ca^2+ + 2Cl^-
This means that for every mole of calcium chloride, there are 2 moles of chloride ions.
Therefore in [50/111] mole of calcium chloride, there are 2 x [50/111] moles

Hence the concentration of chloride ions is 2x50/111 M

Note: if you worked out the above problem in this way, you do not even need the first sentence of the problem. Try working out your mole concept ...not by memorising formuli ...which is an old-fashioned way of working out the problem.

Try reasoning. That's what I teach my students in Singapore.
Discuss with your teacher the above working and I will be more than happy to correspond with you regarding what he has to say. Just write me at [e-mail]

Answer 3

% = 23.7x100/ (23.7+ 375) = 5.94%

M = (d x 10 x %/MM ) x2mol Cl/1mol salt

= 1.05 x 10 x 5.94/ 110.9 =0.562M in CaCl2
x 2
= 1.12M in chloride ion

Answer 4

I've answered the prelab and calculated the same thing as Lexi R. It's correct. Also, good luck on your midterm, if you're in the same class!