5

The technician began his analysis by preparing a Mn2t stock solution. To prepare the Mn2+ stock solution_ 0.6252 g of MnSO4"HzO (MM 169.01 g/mol) was weighed o...

Question

The technician began his analysis by preparing a Mn2t stock solution. To prepare the Mn2+ stock solution_ 0.6252 g of MnSO4"HzO (MM 169.01 g/mol) was weighed on the analytical balance. The salt Ijas dissolved in 50 mL of 2M HNO3 and then analytically transferred to a 250.0 mL volumetric flask and diluted to the mark with 0.18 M HNO3: Calculate the concentration of the Mn2+ stock solution in ppm prepared from the salt 480.8 ppm50.81ppm203.2 ppm2501 ppm812.9 ppm

The technician began his analysis by preparing a Mn2t stock solution. To prepare the Mn2+ stock solution_ 0.6252 g of MnSO4"HzO (MM 169.01 g/mol) was weighed on the analytical balance. The salt Ijas dissolved in 50 mL of 2M HNO3 and then analytically transferred to a 250.0 mL volumetric flask and diluted to the mark with 0.18 M HNO3: Calculate the concentration of the Mn2+ stock solution in ppm prepared from the salt 480.8 ppm 50.81ppm 203.2 ppm 2501 ppm 812.9 ppm



Answers

A stock solution containing $\mathrm{Mn}^{2+}$ ions was prepared by dissolving 1.584 g pure manganese metal in nitric acid and diluting to a final volume of 1.000 L. The following solutions were
then prepared by dilution:
For solution A, 50.00 mL of stock solution was diluted to 1000.0 mL.
For solution B, 10.00 mL of solution A was diluted to 250.0 mL.
For solution C, 10.00 mL of solution B was diluted to 500.0 mL.
Calculate the concentrations of the stock solution and solutions A, B, and C.

Chapter six Problem 44 is a serial dilution problem. This problem tells us about how an original stock solution of manganese was created and how it was diluted into Solution. A, B and C were asked to find the concentration of all of these solutions. So let's start with our stock solution. We're told that we have 1.584 grams of manganese, and we're told that this is dissolved in a final volume of one leader. From this information, we can find the concentration or the polarity of the solution. Remember that polarity is moles per leader, so we'll need to convert these grams into moles. And we do that with a Mohler bass that we get from the periodic table. The periodic table tells us that one bowl manganese is equivalent to 54.9 grams. Therefore, we have 0.28 bowls because this is in a volume of one leader. We simply divide this by one, and that gives us our mill arat E of 0.28 more. Next we're told that this solution is diluted to solution A by putting 50 mils of that solution and they're looting it up to 100 MIT 1000 mills to find with this diluted concentration now is we will use the M one v one equals M two V two equation I m one and V one r. The polarity and volume of our initial solution and M two and V two are the polarity and volume of our final solution. Because we're looking for the polarity of our final solution, we can rewrite this equation as m one V one over V two equals M two. Now let's plug in what we know and one was 0.28 more. The one is five 50 mills and V two is 1000. We can keep the units as mills here because they'll cancel out anyway. But you could convert toe leaders if you want a toe solving this out. This gives us 0.14 Moeller for solution A will repeat the process for solution beat for solution be. We're told that we start We took 10 mils of solution A and diluted it to 250 mills. So using the same equation, we'll take the polarity of a times the volume of A and Divide by the final volume, and that gives us the polarity of B as five 0.6 times 10 to the negative five. You could write this out as a decimal if you wanted Teoh, but at this point we have a lot of zeros, so it's easiest to write it in scientific notation. Okay, now, for our last dilution in this serial dilution problem. Her solution. See, we're taking 10 mils of solution A and diluting it to 500 melts. So following the same procedure, we take the polarity of solution be multiply it by the mills of solution be and divide by the mills of our final solution that gives us a final mill. Aren t for solutions, see of 1.12 times 10 to the negative six.

For this problem will be considering some cereal delusions made from a stock solution. And you'll know what I mean when we get to it. So, first of all, we're given the following information. Um, we have a stock solution that was prepared by dissolving 1.584 grams appear Mangga knees metal. I'll give that a solid in nitric gas it. And then we diluted this to a final volume of one 0.0 leaders. So 1000 mill leaders or one leader, 1.0 leaders. Okay. And then we've got three delusions that are made following this and then delusions Air made. So, for instance, we make a and then make a from the stock that we may be from a and we make see from B. So that's how we're going to do this. So the first thing we're gonna do is figure out, um, what the concentration of our stock is. They were going to say how to prepare a than how to prepare, be and how to prepare. See? Not to bet. So our first order of business will be to figure out our polarity polarity of the stock, which we will do on our next page. I'm gonna check. Make sure my cameras on it. It is. So the stock. Now, um, we're gonna have to recognize a couple of things. 1.584 grands grams of magnesite. Minga knees. Um, if I figure out how many moles of manganese this is, I know that moles of Mayangon ease will be equal to the number of moles of manganese nitrate. And we can tell from the chemical formula that if I know how many miles of manganese I have, I know how many moles of manganese nitrate I have. So for our first order of business, let's figure out bowls of manatees and the molar mass of manganese is 54 0.94 grams per mole. I just looked it up and I get to four Sig fig 0.28 83 malls. There's my first thing that I have, And then if you recall, we dissolved this in there in the nitric acid and then we diluted it toe one leader. So my mill aren t is gonna be really simple. Zero point 02883 in 1.0 leaders equals 0.2883 Moller. That's the Moeller polarity of the stock solution. Okay, easy, Easy. Um, next What do I need to do next? Next? We're going to do solution A and let's go to purple for solution A and were given the following information about solution. A solution? A is I gotta find where I am. Here, There it is. We take 50 mills, 50.0 for six pigs of the stock. And we know the polarity of the stock. And we dilute this to 1000 mill leaders. So we're going to use, um, polarity and we'll go the diluted, diluted and polarity of the stock Times volume of the stock and we're gonna be solving for similarity of diluted. So the polarity of the diluted equals Hillary. The Stock Times The volume of the stock divided by the volume of the deluded Let's substitute our values Bloody of our stock is 0.2883 Moeller Volume of the stock waas 10.0 Mill leaders Nope, That was 50. My bad. Let me raise that 50 milliliters. Excuse me for doing that. And 1000 mill leaders do your math on this one, and that equals 0.0 1442 Moller for solution A. You may have a teacher who who's particular about scientific notation and when to use it. Of course, that will be. There's the solution. There's the, um, concentration explain expressed in scientific notation. Now we have two more to dio. So, um for solution be and we'll go ahead and pick this for B. We're told that we have 10 milliliters of a this time we're taking a and we're diluting it to 250 mill leaders. So again, we're going to use the same concentration of the same formula. And this time, um, I'm going to set it up how I usually set it up because I I'm pretty good at rearranging in my head, so I just sort of don't worry about doing that. So we're going to take the polarity of our stock, which was 0.1 for four. To Moller times the volume of our stock equals X times to 50.0 milliliters and just plug your values in. So and I got 5.7 68 times 10 to the minus fifth Moller for solution be their solution be and last but not least solution. See, we're told that we are taking 10 milliliters of be and we're gonna do that. Dilute that to 500 milliliters. So same set up. We're just gonna have Ah, I'm gonna use the same equations I used in the step above. But I'm going to start with my mil arat e of b times How maney mill leaders or be my volume would be It's gonna equal acts times my volume of my delusion get all done. So for this and I got 1.154 times 10 to the minus six polarity for the concentration of solution, see problem solved.

Here we are looking at the mill aridity, which indicates the number of moles of assaulted per liter of solution. And it's one of the most common units used to measure the concentration of a solution. So in this fast example we've got, the polarity Is equal to the mass divided by the atomic mass, multiplied by 1000, divided by the volume in Millilitres. So for example, in the fast one we have five times 10 to the -9. Fight fight 200.59 Multiplied by 1000, divided by one. We got 24.98 parts per billion. And then what we do is repeat this process for the remaining three examples, using the same equation that we have at the top to get the following values 8.37 ppm, Followed by 1 3 3.47 hot per million. And lastly not .2822 parts per million.

Hi there. In this question, we are converting from parts per million or parts per billion depending on the problem to mole arat e eso. It is important to know that ppm or parts per million for a dilute solution would be the same as 1 mg of salute for every leader of solution. Similarly parts per bilion. But now we're talking about billion instead of million. So that's another factor of 10 to the third. This would be equal to one microgram of the salute for every leader of solution. So using this information should allow us to go ahead and convert from parts per million or parts per billion two molar ity. Remember? Of course, Mole Arat e his moles of solute per leader of solution. All right, With all of that information in mind, let's go ahead and get started. Letter A gives us 5.0 parts per billion. So based upon what we have up here, for parts per billion, this five parts per billion would be the same as 5.0 micrograms of the salute. And the salute in this case is H G per leader of solution. Alright, so that gives us a good starting point because we know polarity is moles per leader. So if we can convert the micrograms of Mercury two moles, we will have polarity. So let's go ahead and do that. First thing I want to do is to convert to Grams. So there are one times, 10 to the sixth micrograms of mercury in every gram of mercury. I wanted to convert to grams so that now I can use Mueller Mass looking at the periodic table. There's 200 0.6 g of mercury in every mole of mercury, right? If we check out what units have canceled, micrograms have canceled, grams have canceled. We're left with moles per liter, and those are the exact units that we need. Formal arat e. So I'm going to grab my calculator now and calculate the answer, rounding it to two significant figures because that's what we were given in the problem. I get 2.5 times 10 to the negative eight moles per liter or Mueller. Alright, so that is the general format we are going to follow for each of these problems. Let's move on to let her be and let her be. We have 1.0 parts pavilion of C H C. L three Again, this is parts per built pavilion. So that would be 1.0 micrograms of C h C l three in every leader of solution changing micrograms 2 g one times 10 to the sixth micrograms in every gram of C h c l three. Then I need the molar mass of C H C l three looking at the periodic table. Adding together one carbon, one hydrogen in three chlorine. Give me 1. 19 0.37 grams of C h C l three in every mole. Okay, just like in the previous problem, all of my units will cancel, except for moles per leader, which is more charity. Calculating the answer. I get 8.4 times 10 to the negative ninth Moeller. So these are very dilute solutions. As you might have guessed, when you're talking in terms of one part per billion, let's go on the part C part C, We are bumping it up two parts per million. So these are a little bit more concentrated. This is our first solution. Are this our solution for part c eso using what we had appear for parts per million. It's milligrams per leader. So that would be 10 0.0 milligrams arsenic in every leader of solution. All right, I am going to convert milligrams 2 g. There are 1000 mg or one times 10 to the third milligrams in every gram. Now I need the Moller Mouse for the arsenic. In one mole, there are 74.92 g. Solving this gives me 1.33 times 10 to the negative fourth Mueller. So, as you can see, this is more concentrated than our previous two solutions. All right, Our final one is letter D letter D. We have 0.10 parts per million of DDT. And just so we know, DDT is C 14 h nine c l five. But in the interest of keeping this recording under a half on hour, I am just going to write DDT each time that I need to. But remember that DDT stands for that formula. Alrighty. So we have 0.10 milligrams of DDT in a leader of solution because that is the equality Four parts per million going ahead and converting from milligrams 2 g when times 10 to the third milligrams of DDT is 1 g of DDT, and then we need the molar mass. So we do need to look at the actual formula here 14 times the mass of carbon plus nine times the mass of hydrogen plus five times the mass of chlorine gives me a Mueller Mass of 354 0.46 grams of DDT. And this is the amount in every mole of DDT. All right, One last time. Calculating arm polarity. We get 2.8. Let me fix that a little bit. 2.8 times 10 to the negative seventh Moeller. So this one's less dilute again, which we could have guessed. If we look, we always start with 0.10 parts per million. Eso that is gonna be less dilute and the molar mass is quite large. All right, so, going back, we have our four answers here in terms of polarity for these solutions. Thank you so much for watching


Similar Solved Questions

5 answers
Identile 'struc/uns . tn Vut photoon Ihe rizhKiuoFunct |Rrchons it oPeraleVSM neerb Ilebo-i crganizaton 1+ Itr mTIuS Svaenkr Irrt :surlaceof twbrsmn t Il InsldMIeFoad for uurlf Wab ltr siles thal dscribe kiday transplnt ccarn nndeL indtogl "sutnnurize Lwir conlvt:ciale-is treatment_ List twa sibes04t7/ /nt /VRnetda t
Identile 'struc/uns . tn Vut photoon Ihe rizh Kiuo Funct | Rrc hons it oPerale VSM neerb Ilebo-i crganizaton 1+ Itr mTIuS Sva enkr Irrt :surlaceof twbrsmn t Il Insld MIe Foad for uurlf Wab ltr siles thal dscribe kiday transplnt ccarn nndeL indtogl "sutnnurize Lwir conlvt: ciale-is treatmen...
5 answers
6.(A) Potential energy stored in a compressed spring is 3 Joules_ The spring constant is k-15000 Nlm (a) Calculate the magnitude of the force in this spring:
6. (A) Potential energy stored in a compressed spring is 3 Joules_ The spring constant is k-15000 Nlm (a) Calculate the magnitude of the force in this spring:...
5 answers
7) Arrange = Ihese compounds butylamine in order = of incrcasing NOH baskcity: 4<I < IV aniline <II < 4< I -butandl I<II<IV < I W<<I < h81 What f Ihe product of Ine following reactton -NaOCHZCH}Co,ChCH,2CH;CH,I H,o" heatCo,CH CH ChCH |co-Ch_cH; CHzCH;CH}Pabr
7) Arrange = Ihese compounds butylamine in order = of incrcasing NOH baskcity: 4<I < IV aniline <II < 4< I -butandl I<II<IV < I W<<I < h 81 What f Ihe product of Ine following reactton - NaOCHZCH} Co,ChCH, 2CH;CH,I H,o" heat Co,CH CH ChCH | co-Ch_cH; CHz CH; CH}...
5 answers
Syppose C is the triangle with vertices (2,0,0),(0,4,0) ganakoerdc above_ oriented counterclockwise as viewed fromUse Stokes' Theorem to evaluate (3y,Vy,z ) dr
Syppose C is the triangle with vertices (2,0,0),(0,4,0) ganakoerdc above_ oriented counterclockwise as viewed from Use Stokes' Theorem to evaluate (3y,Vy,z ) dr...
5 answers
(a). Solve the system of linear congruencies using Chinese remainder theorem x =l1modulo 21(5 Marks)x = 6 modulo 19x =7modulo 113(6). Find the inverse of A =~4 -33Also, find 6A-1(5 Marks)
(a). Solve the system of linear congruencies using Chinese remainder theorem x =l1modulo 21 (5 Marks) x = 6 modulo 19 x =7modulo 11 3 (6). Find the inverse of A = ~4 -3 3 Also, find 6A-1 (5 Marks)...
5 answers
If a ray of light enters the prism hypotenuse-face at 0' incident angle, then the minimum index of refraction of the prism for which the ray totally internally reflects at each of the two sides making the right angle will be given by:air459n = ?450n = 1.326n = 1.273n =1.150n = 1.414n = 1.351
If a ray of light enters the prism hypotenuse-face at 0' incident angle, then the minimum index of refraction of the prism for which the ray totally internally reflects at each of the two sides making the right angle will be given by: air 459 n = ? 450 n = 1.326 n = 1.273 n =1.150 n = 1.414 n =...
5 answers
Evaluate lha follodng integral using the Fundamantal Theorem Calculus Skotch the graph 0/ (ho Intogrand and stde Ihe roglon whose nel drua you have found. Ll-eJLl-H-0 Choose the colrect graph bolox
Evaluate lha follodng integral using the Fundamantal Theorem Calculus Skotch the graph 0/ (ho Intogrand and stde Ihe roglon whose nel drua you have found. Ll-eJ Ll-H-0 Choose the colrect graph bolox...
5 answers
Solve over [0.23];Co8 2*~Cosr =0B. 2cog % col rFind the domain of each function:13 f()= V3x? Srg(x) = arccos(3 + 2x)
Solve over [0.23]; Co8 2*~Cosr =0 B. 2cog % col r Find the domain of each function: 13 f()= V3x? Sr g(x) = arccos(3 + 2x)...
1 answers
Perform the indicated operations, expressing answers in simplest form with rationalized denominators. $$\sqrt{\frac{6}{7}} \sqrt{\frac{2}{3}}$$
Perform the indicated operations, expressing answers in simplest form with rationalized denominators. $$\sqrt{\frac{6}{7}} \sqrt{\frac{2}{3}}$$...
5 answers
Graph the solution of each equation on a number line.$$-6 r=-18$$
Graph the solution of each equation on a number line. $$-6 r=-18$$...
5 answers
Find the area between $y=e^{x}$ and $y=e^{2 x}$ over $[0,1]$
Find the area between $y=e^{x}$ and $y=e^{2 x}$ over $[0,1]$...
5 answers
Evaluate the triple integral f f [2z6+y)av, where G is the solid hemisphere x2+9+z$1,220.1 8 21 3
Evaluate the triple integral f f [2z6+y)av, where G is the solid hemisphere x2+9+z$1,220. 1 8 2 1 3...
5 answers
Which of the following statements is FALSE about the activation of insulin receptor?Binding of two insulin molecules Auto phosphorylation Dimerization after the binding of the ligand Activation of Ras/MAP kinase pathway Activation of PI-3 kinase pathway
Which of the following statements is FALSE about the activation of insulin receptor? Binding of two insulin molecules Auto phosphorylation Dimerization after the binding of the ligand Activation of Ras/MAP kinase pathway Activation of PI-3 kinase pathway...
5 answers
Fmban%utoobieck ul difletrnl cullcbjco Tnne klcunernfchicdE CMcci L #hich #4 mnIluxlly ueaeiling * IQ Ins E dellkcceed Jud" f10t ILs errinal durccilon; ntul (eci #elcke intnt En Lbo oninnal Hunctonu4Fatul Ilc celordyoalch cblcc nftctIlie eublhkon(J Cakulale Io coxreyEatung tha colliston;
Fmban% utoobieck ul difletrnl cullcbjco Tnne klcunernfchicdE CMcci L #hich #4 mnIluxlly ueaeiling * IQ Ins E dellkcceed Jud" f10t ILs errinal durccilon; ntul (eci #elcke intnt En Lbo oninnal Hunctonu4 Fatul Ilc celordyoalch cblcc nftctIlie eublhkon (J Cakulale Io coxrey Eatung tha colliston;...
5 answers
Which of the following substrates will have the slowest S,2 reaction?Ow 00
Which of the following substrates will have the slowest S,2 reaction? Ow 0 0...
5 answers
Y= Xy-2x-7, and y= 8has an area that can be expressed as("(ev+?+R)aywhere a,B,P,Q,and R are constants with & < BMatch the constants with their correct values Partial credit will be given for each correct match: No points will be deducted for incorrect matches2-4VII: 2 VIII: 3
y= Xy-2x-7, and y= 8 has an area that can be expressed as ("(ev+?+R)ay where a,B,P,Q,and R are constants with & < B Match the constants with their correct values Partial credit will be given for each correct match: No points will be deducted for incorrect matches 2 -4 VII: 2 VIII: 3...

-- 0.023683--