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Specific conductance of $0.01 mathrm{~N}$ solution of an electrolyte is $0.00419 mathrm{mho} mathrm{cm}^{-1}$. The equivalent conductance of this solution will be(a...

Question

Specific conductance of $0.01 mathrm{~N}$ solution of an electrolyte is $0.00419 mathrm{mho} mathrm{cm}^{-1}$. The equivalent conductance of this solution will be(a) $4.19 mathrm{mho} mathrm{cm}^{2}$(b) $419 mathrm{mh} mathrm{cm}^{2}$(c) $0.0419 mathrm{mho} mathrm{cm}^{2}$(d) $0.209 mathrm{mho} mathrm{cm}^{2}$

Specific conductance of $0.01 mathrm{~N}$ solution of an electrolyte is $0.00419 mathrm{mho} mathrm{cm}^{-1}$. The equivalent conductance of this solution will be (a) $4.19 mathrm{mho} mathrm{cm}^{2}$ (b) $419 mathrm{mh} mathrm{cm}^{2}$ (c) $0.0419 mathrm{mho} mathrm{cm}^{2}$ (d) $0.209 mathrm{mho} mathrm{cm}^{2}$



Answers

Specific conductance of $0.01 \mathrm{~N}$ solution of an electrolyte is $0.00419 \mathrm{mho} \mathrm{cm}^{-1}$. The equivalent conductance of this solution will be (a) $4.19 \mathrm{mho} \mathrm{cm}^{2}$ (b) $419 \mathrm{mh} \mathrm{cm}^{2}$ (c) $0.0419 \mathrm{mho} \mathrm{cm}^{2}$ (d) $0.209 \mathrm{mho} \mathrm{cm}^{2}$

Morality is calculated as moles divided by leaders. And so if we have 1.457 moles of K C L in 1.5 liters of solution Yeah. Okay. Armel Arat, ease 0.9 seven. If we have 0.515 g of H two s 04 we first have to convert that two moles and the molar mass of H two s 04 is 98.78 grams and we get 0.5 moles. If that's in one leader of solution, that gives us one. I'm sorry. Point 005 Mueller, If we have 20.54 g a l and 03 three convert that to moles. First using the Mueller maths and the molar mass of that is to 12.994 and you get point 10 Mole, once you're a mole divided by 1.575 leaders. Yeah, equals Why 06 Mohler, If we have two point 0.5653 moles of B R. Two in 10 mL of solution. You have to convert that toe leaders So 10 divided by 1000 gives us 0.1 leaders. Did you get 0.5653? But by a 0.1 leaders, That gives us 0.56 53 Mueller.

We're continuing to look at reactions and increase solutions here on DSO For our first reaction, we have Casey l that is titrate id with a g n 03 So our case, er material does not get dissociated do through electrolytes strength off the a g n 03 material and therefore, when Casey L. Is on its own present in the solution, it will have a high conductivity because of its Ionic character. However, when we add a john 03 only some fraction of KCR gets dissociated and remaining in the same phase, so this will decrease the conductivity off the solution. And so the reaction that is listed as a best represented using graph one. Now if we take a look at the next run for the second reaction, Ko age is a strong base added to H air hydrogen fluoride. This has a low acidic nature, so on HF is present alone and the solution will have a low conductivity to to the strong hydrogen bonds. However, one coaches added, that solution is quickly dissociated into K plus. No h minus was then combined with the H plus of hydrogen fluoride, and it increases the mobility off all containing irons, which will increase the conductivity of the solution. So therefore, be fast represented by graft to finally, in the third, reaction of conductivity of the solution gradually increases due to the removal off the glass immobile be a two plus ions on s 04 to minus by the highly mobile n A plus and C l minus Having a greater number off moles. It's what we see. A currency, There will be a two clubs Got to see out minus two and a yeah ad s for Tu minus in equilibrium with B I best for two and a plus to see out minor. So we have is no 0.1 Moloch not point to not point to. Don't 0.1. Not quite one not point to and not point to

Hello Today we're doing problem 36 from chapter four. They gave us Equus Solutions and they were tested with light bulb conductive ity apparatus and which result dark, dim or bright Do you expect from each? So they gave us potassium chloride, methanol and acetic acid. I wrote each of the Myra's a the thief. So first thing we want to do is what would turn on this light cold. We want to make it conductive. And conductivity is based on bionic ability. Right? So if we have ions floating in here, it's gonna make this turn bright. It's gonna turn on. And I wrote our first example here we have K plus and C minus makes clear here feel minus. So those ions fully dissociating, which we know Casey l will fillied associate is going to mean that we're gonna have a bright bull. We're gonna have high connectivity because of those ions. And that bull was going to be bright looking at EMI. Ohh. Now we have a metal with an O age attached. That's what methanol is. We know that it's not gonna associate. There are no ions in it for it to disassociate. It does not have any conductive ability, so it's going to be a dark holes looking at our last one. We have acetic acid. You should know that this is a weak acid. Well, I said, and we know that week assets don't fully dissociate. They only partially associate with their aisles. So we're gonna have some ch three c o. H is in the solution some h plus and some conjugated acid in the solution. So because it's partially associating, it's going to be it's not gonna be as bright as the K C L, which fully associates, but it's gonna

Hello And this question we are asked if a, uh of a solution with a certain compound will create Well, um, create an electrical current in order to support the lighting up of a light pole. And our first question, we are asked, if a solution of Hydrofluoric acid onda and this we know that a Jeff has a small as a small the student association constant and therefore doesn't associate so much. But even though it does this it even though just so see, it's a little it still associates enough that we expect a dim light in our next question. And the H Plus an F minus are able to sustain an electrical a small electrical card. For next question, we have a compound of sodium chloride, 0.1 Moeller on and okay, C l and this one n hcl is a straw. It just has a really it was completely associates. And therefore we know that it has, and what's called is able Teoh have a pass on an electrical current, and we expect therefore to be a bright light in our next question, we have a solution off. Uh, glucose and glucose is not is not an electrolyte and therefore and does not associate in water. So glucose does not associate water and therefore is not able to conduct an electric current and therefore there is no light.


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