Describe how you would prepare each of the following aqueous solutions, starting with solid KBr: (d) a 0.150 M solution of KBr that contains just enough KBr to precipitate 16.0 g of AgBr from a solution containing 0.480 mol of AgNO3.

Video transcript

Hello everyone. Today. We are being asked to describe the preparation process that's needed to make .17 moller of rubidium hydroxide solution from solid rubidium hydroxide to precipitate 12g of silver hydroxide from a solution with .125 moles of silver hyper chlorate. And so the first thing we want to do is we want to start with what we're given. And so we have our 12 g of silver hydroxide. We must somehow convert this into molds of rubidium hydroxide. So we're going to do, we're going to multiply this by the molar mass. We're gonna say one mole of silver hydroxide is equal 224.88 g of silver hydroxide. And then we're going to perform a multiple ratio. We're going to say that for every one mole of silver hydroxide we have one mole of rubidium hydroxide. Our units are going to cancel out And we're going to be left with 0.09609 mol of rubidium hydroxide. We must then take that rubidium hydroxide that we sold for. So we have 0. moles of rubidium hydroxide. And we must convert this into leaders. So what we can do is we can multiply By the inverse polarity that we have. So you have a polarity of rubidium hydroxide of one leader equaling 0.170 moles of rubidium hydroxide. We multiplied by the inverse morality. So that our units could cancel and we're left with 0.565 L. And so now we must take our original moles once again. So we're going to take our 0.09609 moles of rubidium hydroxide. And now we want to find how much that is in g. And so to do that, we must multiply by its molar mass and we do the inverse as well so that our units can cancel out. So we'll say one mole of rubidium hydroxide is equal to 102.475g of rubidium hydroxide. Once our units finally canceled, We're going to be left with 9. 468 g of rubidium hydroxide. And so our final answer is we're going to dissolve This 9.8468 g of sodium hydroxide and a small amount of water and then dilute To our .565 L. And with that we have our final answer. I hope this helped. And until next time.