How many 9rams ot solute arc picsenimL 0 1J3 CoSO4?(D) I{ 1,50(NH4)SO4 Is dissolved enough WuteiIarm 250. ML ef solutlon, whot = the molarity 0 the solutlon?(c) How...

How many grams of solute are needed to prepare each of the following solutions? a. $1.000 \mathrm{L}$ of $0.200 M \mathrm{NaCl}$ b. $250.0 \mathrm{mL}$ of $0.125 M \mathrm{CuSO}_{4}$ c. $500.0 \mathrm{mL}$ of $0.400 M \mathrm{CH}_{3} \mathrm{OH}$

When one substance is dissolved into another, we form a solution and the solution is a homogeneous mixture made up of a solid dissolved into a solvent. So the first thing we're looking to do here is determine the mold K C L, Which is the mass divided by the Molar Mass. We end up with 9.190 moles. And then to calculate the polarity, we take our moles divided by the volume 9.250 l To get 9.7619 Mola You want around that 2, 3 significant figures. We get 9.762. And then we just repeat the exact same calculation for the remainder as well. Run through one more so fastly calculating the malls of K2, CR 04 Mass divided by the Molar Mass. We get not point not to 62 moles. What we then do is divide this by the volume 9.150 L. To get the polarity, not .1743 Mola. If you want to, three significant figures, we get 9.174 and so there are two remaining examples that we have. We end up with 9.9126 And not 167 molars, doing the exact same calculation while we fast work out the moles, and then we can calculate the polarity using the volume.

A few calculations to make our way through. So we have a lot of given data. So in the part A. So we can start with is calculating similarity which is the moles of solid divided by the volume of solution of liters. So we can write all the above expression for the moles of solute is equal to the polarity multiplied by the volume of solution and liters that is equal to 9.736 mola multiply that by not point not 157 liters. What we get is not point not 110 moles. So because each model of K two cr 207 has a massive 294.19 grams, the required number of grams of K two cl two oh seven is as follows. The grams of solid is equal to not point not 110 moles multiply that by 294.19 g. Put mole. We get the value of 3.24 g for the fast part. So quite a few cynical calculations. The next. What we have is again, a lot of given data. The first thing we can do is find the number of moles of NH 42 S 04 So we have the malls is equal to the mass divided by the molar mass. So that is equal to 14.0 grounds divided by not 0.10 divided by 1 30. Sorry, 132.14 g per mole. We got not 0.1059 moles. So we know that the polarity is equal to the moles divided by the volume. So we have not 0.1059 moles divided by not 0.250 liters. We get not 0.4 to four mola bump. So next again we just have a series of data that's presented to us. We can calculate the malls of see us A four. So in part C we have the moles that's equal to 3.65 g divided by 15961 g per mole. So you have the weight divided by the molecular weight gives us not point not to to eight moles. And then following this, we know that the polarity is equal to the malls of the soviet divided by the volume of the solution in liters. And what we're able to do is rearrange the volume. So what we have is the moles which is not point not to to eight moles. Try that by the polarity, which is not point not 455 lola. And what we get is 501 millimeters.

When one substance dissolved into another, a solution is formed and a solution is identified as a homogeneous mixture. That is made up of two things. Firstly, we have a solid that is dissolved into our second thing, the solvent. So the first thing we're looking to do and part is calculate the moles of an age four cl, But just 3.00 moles. And so we can calculate the mass of n H four cl. Now that we have the moles. So we take three malls, multiplied by 53.49 grams per mole. And what we get is 1 60.47 g. So therefore we need 100 and 60 g of NH four cl and then we do the exact same thing with the remain ear's. So we have the moles. Okay, Ceo, So that's equal to 9.75 moles per liter, multiplied by 1.5 L. to get 1.125 moles, multiplied by that by the molecular weight from the periodic table, 74.55 g per mole. Then we get the mass Of KC. L, which is 83.86 g. If you want around that for your significant figures, we've got 83.9. Finally, one more example, we start off with the moles again, so moles of Any two s. 04, so what we have is Not .35. So we take 9.35 zero mph. Multiplied by 9.15 L to get not point not 5-5 moles multiply that by the molecular weight from the periodic table, Malema's one for 2.4 g per mole to get the mass of any two S. 04 That is 7.4517 g. And then for rounding for your significant figures, we have 7.46 g.

Guys were doomed. Problem 96 Chapter 11 20 in chemistry 2017. So a 20170.500 gram sample of a compromise dissolved Nut wards make 100 millions a solution. The solution has an osmotic pressure of 2.50 a. T. M. At 25 degrees Celsius if each one like you. So you're associating too Particles in this solvent was the Moler massive? That's all you We're gonna use this formula Pi is equal to I m r t where eyes, the pressure and miss concentration are the ideal gas constant and t is temp temperature. And Kelvin, um, I in this case because each because all each molecule so it dissolves into particles that's going to our eyes gonna be equal to two because I is the ratio of the number of particles over the number off molecules I saw you dissolved. So now let me solve Molecular great member M is equal to end over B and and is equal to em over and w So we plug everything in. So that means high is going to equal to I am our key over the MW. So that so then we so for MW That's gonna be MW is equal to I m r t over be high So then we plug everything into two times zero point 500 grands times is your point is your eight to 06 leaders Adm her more Kelvin were ats 25 degrees Celsius. So you had to 73 to that. That creates to 98 Kelvin over 100 mil leader. So zero point 100 leaders and multiply that by 2.50 p. M. So m w people like everything into two times 20.59 point 08 to 6 time soon indeed, No, by 0.1 times 2.5. That's gonna give us a molecular weight of 90 seven point a grams per month.