imagine-that-you-have-500-mathrmml-of-020-mathrmm-sitrate-solution-how-many-milliliters-of-050-mathr

Name: imagine that you have 500 mathrmml of 020 mathrmm sitrate solution how many milliliters of 050 mathr
Uploaded: 2020-10-18
Duration: 3 min 36 s
Channel: Numerade
Description: Imagine that you have 500 $\mathrm{mL}$ of 0.20 $\mathrm{M}$ sitrate solution. How many milliliters of 0.50 $\mathrm{M}$ sodium chloride solution should you add to remove all the silver ions from the solution?

Question

Imagine that you have 500 $\mathrm{mL}$ of 0.20 $\mathrm{M}$ sitrate solution. How many milliliters of 0.50 $\mathrm{M}$ sodium chloride solution should you add to remove all the silver ions from the solution?

Vishal Sharma · Accepted Answer

so in problem six, we have given 500 ml off zero point 20 Moeller silver nitrate solution, and what they ask is how many milliliters off zero point 50 Mueller and the solution should you need to remove all the silver ions from these solutions? So that means we have to find out the volume or for initial volume off NSL in militar. So first we see the reaction that can take place between the silver nitrate, echo silver nitrate and eco sodium chloride. So this is the reaction. So this is the silver nitrate echo silver nitrate. When this reacts with the echo sodium chloride, then it will form solid silver chloride along with echo sodium nitrate. So, like this is thes simple balanced chemical equation for this process. Okay, so that means we have given this information This silly echo silver nitrate is 500 ml and it&#x27;s 0.20 Mueller solution and this its volume. We have to find out volume offense, yell, and this is given 0.2 0.50 Mueller and Israel solution. So this is the 0.50 Moeller and a single solution so Basically what we can do is we, uh we can this, uh, morality can apply this majority question. So morality off silver nitrate. So this is the majority of silver nitrate that is given in the question, and we can multiply this city with the volume off Sylvan, a trade volume off silver nitrate solution. And this will equal to the because they combining the one treasure one mole ratio. So this is equal to the majority off this NCL solution multiplied by volume off any single solution volume off any seal solution. Okay, so we can assume that if everyone is the majority off silver nitrate, that is 0.20 Moeller and we one is the volume off silver nitrate. That is 500 ml. So this is I can use for silver nitrate and the majority off sodium chloride. 0.50 Mueller. And the volume that is young for sodium chloride is veto that we have to find out. This is for NSCN. So what we can do is we can put the value in this equation. So we get 0.20 and 2 500 which is equal to 0.50 and to veto. So this 502 0.2 we get 100 which is equal to 0.5 and two. We too. So we to value is equal 200 y 0.5. So on solving this we get 200 ml so that when the volume offer initial solution. So the volume off this n HCL solution in Emmel, which is a call to 200 m l. So this is our answer that 200 ml a volume open ASEAN solution.

David Collins · Answer

it&#x27;s helpful when answering this question to first right the balance chemical reaction so that we see that there is a one toe, one mole relationship between sodium chloride and silver nitrate. We can then take the 25 mL of silver nitrate that is going to react and convert it to leaders, and we get 0.0 to 50 leaders Silver nitrate. Once we have the leader silver nitrate, we can convert it to mold silver nitrate using the polarity silver nitrate and mold silver nitrate to mold sodium chloride from our balanced chemical reaction and then mold sodium chloride to leaders of the sodium chloride solution, and we get 0.20090 leaders or 9 mL.

William Mills · Answer

His question says that you need to make 150 milliliters of a 0.1 Moeller an A C l solution. It says that you have solid any C l and your lab partner has a 2.5 Moeller an a c l solution. And then it asks us to explain how you each make the 0.1 Mueller en hcl solution. So right here we need 150 mils of a 0.1 more an a C l solution. You have solid. And a c l. What you&#x27;re gonna do is dissolved the solid and water. You just need to figure out how much solid you&#x27;ll need. So start with the concentration is your appoint one moons her leader times number of leaders. I&#x27;m just gonna convert 150 milliliters two leaders by dividing by 1000. That&#x27;s 10000.15 leaders. And then I&#x27;m gonna multiply this by the molar mass of sodium chloride, which is 58 0.44 grams per mole. And if it is correctly, all the unit should cancel out moles here, cancels out with moles here, leaders cancels out with leaders. That just leaves us grams, which is what I want to know is the massive sodium chloride that I&#x27;ll need. If you multiply this out, we get zero point eight eat grams sodium chloride, and then he would simply dissolve this amount said in chloride and water and have the final volume be 150 milliliters. For your lab partner. They have a 2.5 moller and a C L solution. Instead of developing a solid will be diluting their more concentrated stock. And so we can use the equation. See one me one equals C to be, too. Which is to say that the initial concentration times the initial volume equals the final concentration times the final volume. The initial concentration of your partners stock solution is 2.5 Moeller me one or how much of that solution don&#x27;t need is what we&#x27;re gonna solve for the concentration they want to make it at the end is 0.1 more and volume is 150 mils. If you divide 2.5 over for V one, you would see that it equals six mills and its middle leaders because male leaders was the units we used for this volume, said six milliliters of 2.5 more. Where and a CEO. Now we need to figure out how much water we need to add to that. You could just say Bring it to 150 as you do in the lab, but we can also just calculate it. It&#x27;s the volume of each to O is going to equal the total or final volume 150 mils minus the volume of any C. L. You&#x27;ll be adding just six mils. That makes the final volume of water needed 144 milliliters of water. To recap for you with your solid and a c L U will dissolve 0.8 grams and bring the volume to 150 or add 100 50 mils. Because it&#x27;s pretty small. Take a pretty small amount of space if your lab partner, since they have a concentrated in a cell, any C l solution the dilute six mils of that solution 2 150 mils by adding 144 miles of water

Iva Lazarova · Answer

for this problem, we want to solve how to create a 0.1 Moeller solution that is only 150 milliliters, both from solid and A C. L. And from a stock solution. Let&#x27;s start with the solid first, nor to figure out how many grams of the sodium chloride we need. We need to figure out how many moles we need. Now we know we have desired volume of 150 millimeters, which is about 0.15 leaders. So we&#x27;re gonna take our final volume. Multiply it by the polarity. 0.10 moves many CEO over one leader of solution That gives us 0.15 Moles. Sodium chloride Need it now we&#x27;re gonna take this. Well, why? By the molar mass of sodium chloride and find out how many grams that is that we&#x27;re using. So 0.15 walls of any seal times every mole oven. A CEO has 58.44 frames. You find the 54 4 by adding the molar mass of sodium in the molar mass of chloride that&#x27;s gonna get us about 0.88 grams of sodium, sodium chlorate. So we&#x27;re gonna take 0.88 grams of sodium chloride, mix it with or rather, added water to that up to the one leader mark and just let everything dissolved. Now, for the&#x27;s stock solution, we start with a concentration of 0.25 more sodium chloride. Want to dilute it down to 0.1 Moeller. In order to do this, we need to figure out how much of the volume of the stock solution we want to put in to make 150 milliliters of the desired concentration. So we&#x27;re gonna use the C one you on equal see two feet to formula R C. One of hers concentration is the original stock solutions to your 10.5 moller times. We don&#x27;t know our volume, so I&#x27;m just gonna leave that as B one equals. Well, our second concentration is gonna be our desired concentration. So that&#x27;s going to be zero point one more times per second volume. It&#x27;s 150 milliliters, which is 0.150 lead. You have to remember to convert everything until leaders from working with this equation. So we have 0.25 b. The one equals 0.15 When you divide, those can be wonderful, your 0.6 leaders, which is 60 milliliters. So now we know we want 60 milliliters of original solution. But how much water do we add to that 60 milliliters? Well, let&#x27;s go over to this side. We know we want a total of 150 milliliters. 60 of that is going to be our original solution, which means we are going to have 90 milliliters of water to add to get it up to are 150 milliliters. So what you&#x27;re going to do is you mix 60 milliliters of stock solution and 90 milliliters of water, and that will get you your desired concentration of William.

Freddie Montague · Answer

for this question we have to solve for the volume of 0.10 Moeller any to solution that would precipitate all the nickel. Um, from the n i c l to solution. So I&#x27;ve re written the balanced equation that were given and in most we&#x27;re gonna write out, um, all the values we have. So the volume of this and I seal too is equal to 25 milliliters, and we also have the concentration is you&#x27;re applying to zero molars and other pieces of information that we have is the concentration of this any two s and that is equal to 0.10 molars. And this is what we&#x27;re trying to find the volume of. We don&#x27;t know the volume of this. We&#x27;re trying to find the volume of this. I would precipitate all of this and I asked, So we can start by noticing that in this balanced equation stoic yama tree of the three species that we&#x27;re interested in he is one toe, one toe one, but also to hear in this and sodium chloride. But we aren&#x27;t looking at this one. This is irrelevant for this problem. So there is a 1 to 1 to one ratio. So that means the number of moles of thes of this a nice eel to is gonna be the same as the number of moles of this, any two s and so the same number of moles will be required. Um will be present in the nickel in the n i s so we can start by solving for the moles of this. And I see all too, since we have the most information. So moles is equal to concentration times, William. So this is zero point 20 molars and the volume is in milliliters. We have to convert this until leaders to use this equation. So we&#x27;re gonna divide the 25 5000 being at 0.25 leaders, and we divide by 1000 because there are 1000 middle leaders in one leader and this keeps us 0.5 nos and again, this is the ratio between the two reactions is 1 to 1. We also know that the moles of this I need to us it&#x27;s also going to be 0.5 And using the same equation is before an equal C times T you can rearrange it and sulk with volume of the any to us, and that is equal to moles over concentration Normal 0.5 and our concentration is zero plane 10 molars. This gives us 0.5 leaders and we multiply this by 1000 and her answer is 50 middle leaders.

Dan Ni · Answer

So in this problem, we prepare this chloride solution and we want to precipitate out all the chloride by adding silver nitrate, which would react like this and form silver chloride and make sure so we know from the store ideology of this reaction that the moles of the chloride ion in solution equals the moles of silver nitrate that we need to add. Right? So if we want to find out how many, how many moles of silver nitrate we need and we need to find how many moles of chloride or in solution. Yeah, pretty easily just take the polarity of an A. C. L. Times the volume of it to find how many moles of an A C. L. There are. And then there&#x27;s one mole of chloride per mole of an AC. And we can do the same with MG cl two. And if we multiply all these together, You get 0.0948 moles of chloride. Okay, This corresponds to this many moles of AGN 03. Now we&#x27;re told that the polarity of the silver nitrate solution. His 0.20. So 0.20 moles of silver nitrate is one leader, 0.0948 moles of silver nitrate is how many leaders, We can just sulfur that. And this corresponds to 0.492 leaders of silver nitrate that we would need to precipitate all of this chloride in the solution. So that&#x27;s the problem

Problem 6 Easy Difficulty

Imagine that you have 500 $\mathrm{mL}$ of 0.20 $\mathrm{M}$ sitrate solution. How many milliliters of 0.50 $\mathrm{M}$ sodium chloride solution should you add to remove all the silver ions from the solution?

Answer

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