To what volume should you dilute 50.0 mL of a 5.00 M KI solution
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Question

To what volume should you dilute 50.0 mL of a 5.00 M KI solution
so that 25.0 mL of the diluted solution contains 3.05 g of KI?

Accepted Answer

Hey everyone today, we're being asked to find the final volume that a sodium chloride solution was diluted to. Now. Right off the bat as you can see that we're dealing with more clarity and volume. And the question, we need to utilize the formula that M1 V one equals M two V two. Where M&V. Sanford more clarity and volume were one and 2 determined before dilution and after dilution. Now in the question were actually given the initial concentration and volume of the solution. That's right here, 20 ml of three molar sodium chloride solution. So let's go ahead and write that down. We'll write that in blue. So you have and one is equal to three molar 3. molar And V one is equal to 30 million. Now we're asked to find v. two. Me too, but we don't know what M2 is the molar itty of the diluted solution. However, we are given the volume of a part of that diluted solution and the mass of sodium chloride it has. So using this we can actually go ahead, convert the mass to molds and then determine Using that value what is the molar itty of the diluted solution. And that can be done with a series of steps. So let's write this out down here and two is equal to and we'll write this like. So we have 0.29-2 g sodium chloride. So to convert this to moles, we need to multiply by the molar mass and we can do this by adding up their atomic mass units on the periodic table, which gives us a final value to have one mole of n A c l for every 58.44 g of sodium chlorate. However, we're not done just yet. We'll have the molds. But we also need to convert the volume of mL to liters because polarity as we can recall, polarity is equal to malls per leader. So we have 10 0. mL multiplied by a conversion factor of one leader for every 10 to the third milliliters because remember we have 1000 mL or 10 to the third milliliters for every one liter four mL will cancel out and we will be left with a final value for M two of 0.5 Mueller. So this is 0.5 molar. So now we can go ahead and solve for V two. Let's do that down here. So M one V 1 is equal to M two V 2. Therefore V two is equal to M one V one by M two, which is simply three hoops. Wrong too. This is just three Moeller times 20 milliliters divided by 0. Molar, which gives a final value. Let's write this in green. A final value Of 120 ml. Therefore the final volume that the original solution was diluted to is 120. I hope this helps. And I look forward to seeing you all in the next one

To what volume should you dilute 50.0 mL of a 5.00 M KI solution so that 25.0 mL of the diluted solution contains 3.05 g of KI?

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