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2 Using the data you obtained from the graph in the previous problem determine the concentration of cobalt (II) in 1.975 g sample of a substance, which was dissolve...

Question

2 Using the data you obtained from the graph in the previous problem determine the concentration of cobalt (II) in 1.975 g sample of a substance, which was dissolved in water in a 100 mL volumetric flask and then diluted to the mark. For the absorbance of the cobalt solution from the flask a value of 0.55 was obtained. What is the percentage of cobalt in the unknown compound? Atomic mass of Co is 58.933.

2 Using the data you obtained from the graph in the previous problem determine the concentration of cobalt (II) in 1.975 g sample of a substance, which was dissolved in water in a 100 mL volumetric flask and then diluted to the mark. For the absorbance of the cobalt solution from the flask a value of 0.55 was obtained. What is the percentage of cobalt in the unknown compound? Atomic mass of Co is 58.933.



Answers

(a) What is the mass percentage of iodine $\left(\mathrm{I}_{2}\right)$ in a solution containing $0.035 \mathrm{~mol} \mathrm{I}_{2}$ in $115 \mathrm{~g}$ of $\mathrm{CCl}_{4} ?$ (b) Seawater contains $0.0079 \mathrm{~g} \mathrm{Sr}^{2+}$ per kilogram of water. What is the concentration of $\mathrm{Sr}^{2+}$ measured in $\mathrm{ppm} ?$

Problem for the party Mass percent equals two mass of absolute, divided by total mass of solution multiplayer. And it Okay plus we need to calculate mass of soldiers mask of salute Or master lute that is equals two 0.0 35 moles. Oh, I do molecules multiply by 2 53 .8g. i two molecules divided there. One mole. I took only two here we multiply the number of moles of item molecules into its molecular, its molecular bit. That is calls to eight .9 g item molecules. So putting this value and this value in this formula we get Mass% 2 molecules is close to master absolute. We already calculated air mass absolute 8.9 ah 8.9 g. I too divided back 8.9 Grand I too Plus 1 25 graham CCL port Multiply by 100. After calculating we get approximately six point 6% I do. This is our final answer. All party party in part B PPM equals two mass salute, divided by total mass. Absolution Multiply by 100. Sorry, multiplied 10 to the post. six. That is close to after putting the value of mass of salute. There is 0.00 79 grams assad plus two and Total mass solution that is one. Multiply. Work 10 to the power three g as to into 10 to the power six. After solving this, we get approximately seven point nine ppm assad plus to this is our panel and support part of it.

Thank you for joining me again or still in Chapter 13 now onto problem 40. So we're looking at what is the mass percentage of iodine in solution containing 0.3 0.35 moles of ID. I'm in 125 grams of C seal four, which is carbon tetrachloride b. We're looking at something very similar. Sea water contains. We've got grams per kilogram of water and what is the concentrations and ppm? So let's start at the top. Worry about. Be a CZ. Well, it's not a part one part two problem. So let's look at what our mass per cent is. Let's try to do this from memory, see if you can write it down. What is your mass percentage formula? Mass per cent is massive. Your Salyut over mass of solution and solutions made up of two parts. It is both saw you'd and solvent added together. And what time's this by 100 to get rid of their percent sign. All right, so let's see what we have. We were given our mass percentage of iodide, which is very small and was it in mass masses and units of grams. So what we're given to start with with I died was 0.35 moles have I died, too? And the reason there's a two there is because iodine is one of those Die Atomics, one of our seven scaredy cats on our periodic table that requires that they're always in pairs. So looking at our periodic table weaken, find a way to convert this two grams, which is what we need. We're going to need grams of both our sol you and are solvent, even if we're not sure which one. This is as we know we have to go into grams, and we know if we use the periodic table, it's always set upto one mole if I died. So let's check out our periodic table and we find that I had died is 126.9 grams primal. And just remember, we have two of these, so we absolutely have to count it twice. So that's 253 point eight grams in our calculator were just times that by r 0.35 moles. When we find out if we weren't away that I don on the scale, we would have eight point 88 grams of either watch out for We're going to have to watch out later when we finish up our problem for significant figures. But we're not going to take that into account just yet. All right. So let's look in the mass per cent. This is we found our Salyut is always the lower number over solution. So we know that we dump this 8.88 grams of iodide too into 125 grams of carbon tetrachloride Which one's larger? Absolutely by far the carbon tetrachloride. So we know that this is our solids. Lets start setting that up. We have our mass per cent. It's equal to our massive Salyut which we have solved already which is eight point 88 grams of iodide Perfect Now our massive solution Our solution is both of these added together. So it is our I died plus our carbon tetrachloride Sorry to 8.88 grams plus what we don't are our ride on into which is the carbon tetrachloride 125 grams. Remember we needed times it by 100 and this unit never came in there and they'll solve So let's put it into our calculator 8.88 and divide it by our solution, which is 8.88 plus 125. And all together we end up with in the Mount that needs to be times by 100. So if you ended up with six 0.63% then you're absolutely correct. That is the percent mass that we have in part A. And I'll give you just a moment here. If you'd like to write anything down, we're going to switch over two part B of our problem. All right, opening up a new page. We're going to work on that. So we have sea water that contains 0.79 grams of this particular cat eye on which is strong Liam two plus. And that's in every kilogram of water per one kilogram. What is the concentration of stro NNI, um, and P P M's? Can anyone tell me what ppm czar parts per 1,000,000? Some. If we want to convert into P P M's parts per 1,000,000 we're just going to use a conversion factor we're gonna have, and all parts per 1,000,000 are is mass of Salyut over Massa's solution and they gave us this. This is our massive arse. All you'd over the mass of our sea water solution seawater solution. All right, so all we have to do is change it from Graham's two kilograms into parts per 1,000,000 and that it's something we looked at doing around chapter two and all we have to dio is change it to parts per 1,000,000. So we're just times it bye 10 to the sixth Power. So what that looks like here is we have our 0.79 grams. Just kidding that zero and that's over one kilogram and we're going to switch back telegrams into grams. All right, now we've got our massive saw you'd over a massive solution and masses and grams, we like it when they're all in the same units. It makes it easier for us and that'll help us switch into Harry R Times 10 to the sixth switches into parts per 1,000,000 on what we find is multiplied us all in our calculator 0.79 divided by 1000 times one times 10 to the sixth power. And now we are 7.9 parts per 1,000,000. I'm strong IAM two plus a cat eye on. So a trick you can use whenever you're doing this on a test is look at the beginning here and look at the end here the parts per 1,000,000 and you'll notice they're very similar. There always be the same amount because we're only going to be doing factors of 10 groups of 10. So if you find a multiple choice question, that's an 8.1 or a 6.4 in a number that's changed. Our seven and nine are absolutely wrong. Good luck on your test, and you know you'll ace this one.

If we have an absorption graph based on concentration of iron ions, it might look something like this where our concentration is on the X axis and our absorption is on the Y axis. So as our concentration increases are, absorption increases with now the Grady int of this, uh, this graph where the slope of this graph is actually going to represent arm Mueller extinction coefficient is because if you remember, um, slope of a graph equals rides overrun. Then in this case, we have absorption divided by concentration and assuming that our path length is the same the entire time, one centimeter and this is actually going to be equal to our extinction coefficient. So this will be constantly entire time. Now, let's say that we are given an absorption value a zero point to you, and we want to figure out what is the concentration at that absorption. Now, there are two ways you can do this. You could come over to the craft and say, Well, our absorption values approximately right here, and so we can trace that back, and so are concentration. Must be approximately right here. But if your graph doesn't really allow you Thio determine the concentration that easily we can always calculate it. So let's say that we need to find our extinction coefficient first so we can grab two points off this graph and find the slope. So let's use this point, which is 0.0.3 point six on this point, which is one in 2.25 These are approximate values, so we're going to save point are 2.25 and a 0.6, all divided by 1.0 minus 0.3. That's going to give us an extinction coefficient or a slope of approximately two point 36 All right, so if we use this as our extinction coefficient, we know our absorption is point Thio. By the way, the units of this are going to be decimate er's cute per mole centimeter. So if we put these in, we remember our beer's law tells us the concentration of a solution is equal to the absorption divided by the extinction coefficient multiplied by the path like so, if we remember this, we can go ahead and plug in our values. 0.2 divided by 2.36 desi meters cubed from mole centimeter multiplied by a path length of one centimeter, and that gives us a concentration of approximately nine times 10 to the negative. Six moles per decimate er cute. And that's how you would find the concentration based on the absorption value and a given absorption graph.

So trans mittens and absorbent are inversely related, so we can calculate the concentrations of cobalt by taking their corresponding Mahler extinction coefficient values and their wavelengths. So we have C is equal to a over e l so cold, but with a wavelength of 510 under meters, we have concentration is equal to 9.446 Divide that by 36400 multiplied by 9.1 That's equal to 1.22 times 10 to the minus. Five more Podesto meats are cute. Next we have 656 nanometers is equal to not point 446 Do I do that by 1 to 40 times 1.0? That's equal to 3.596 times 10 to the minus four most protesters meters. Next we have a nickel to look at. So we've got nickel with a wavelength of 5 10 nanometers. So we have concentration is equal to 9.3 to 6. Invited by 45 to 0, multiplied by 1.0 the length 4 to 5.905 times 10 to the minus five malls decimates a cubed. Next we have neck all the way, but 656 nanometers concentration is able to not 0.3 to 6 by that by 17500 multiplied by 1.0. That gives us a body of 1.862 times 10 to the minus five malls Podesta Meter Cube.


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