A solid mixture of KCl, KNO3, and Ba1NO322 is
20.92 mass % chlori...

Question

A solid mixture of KCl, KNO3, and Ba1NO322 is
20.92 mass % chlorine, and a 1.000 g sample of the mix- ture in 500.0 mL of aqueous solution at 25 °C has an osmotic pressure of 744.7 mm Hg. What are the mass percents of KCl, KNO3, and Ba1NO322 in the mixture?

Accepted Answer

Hi everyone for this problem. It reads a 1.050 g mixture of sodium chloride, sodium nitrate and magnesium nitrate. That has a 12.41% chlorine. By mass was dissolved in a 510 million liter water at 25°C. The result, A solution was found to have an osmotic pressure of mm of mercury, calculate the percentage mass of sodium chloride, sodium nitrate and magnesium nitrate. So the question that we want to answer here is the mass percent of these three things. Okay. And to do this because we're given the osmotic pressure, let's go ahead and write the equation for osmotic pressure. And that equation is osmotic pressure is equal to M. R. T. Where m is the molar mass or the malaria T. Excuse me, R. Is the gas constant and T. Is temperature. So the first thing that we're going to want to do is we're going to we're going to want to manipulate this formula to get the malaria T. Okay. And then we're going to substitute the values given in the problem. So let's go ahead and manipulate this formula so that we're solving for malaria T. And when we do that, what we're going to get is malaria T. Is equal to osmotic pressure over R. T. So looking at the problem, we're given the osmotic pressure, we're told that it is 745 millimeters of mercury. The gas constant. R is a value. We should know and that value is 0. leaders. atmosphere over mole kelvin. And the temperature is given in the problem to be 25 degrees Celsius. Okay. And we're going to want to convert this temperature to kelvin because our gas constant R. S temperature unit is kelvin. So 25 degrees Celsius plus 273.15 will give us the kelvin temperature and the kelvin temperature is 298 point 15 kelvin. Let's just move this over. Okay, so now we have everything we need and we can go ahead and plug that in so that we can do step one which is solving for the mill arat e. Okay, so what we're going to get is The osmotic pressure which we're told in the problem is 745 mm of mercury. We're going to want to convert this two atmospheres because our gas constant unit for the pressure is an atmosphere. So we're going to multiply this by one atmosphere over 760 millimeters of mercury because that's how we convert from millimeters of mercury to atmosphere. Okay, And this is all over the gas constant R which is 0.8206 liters atmosphere over mole kelvin times the temperature in kelvin, which is 298.15 kelvin. So now we have everything we need. So let's solve for the malaria t When we get this, what we're going when we saw for this, what we're going to get is 0. molar. Okay, So now that we know the mill arat E. Let's go ahead and get the total number of moles from this modularity. Okay. And we're going to get the total number of moles By using the conversion. That is given in the problem where we have 510 milliliters of water and this is per mole ion. All right. And so what we're going to get then is this similarity is our starting point? So we have 0.4006. And remember um Alaric e is moles over leader. So that is moles over leader. And we're going to multiply this by the milliliters of water. But we're going to convert our milliliters to liters. Okay, so let's write this another way. So this makes sense. So let's use our starting point to be the volume of water. Okay, so our starting volume, our starting point for the volume of water. We're going to convert milliliters to liters. Okay, so we're going to move this three units to the left. So we have our starting point is 0. leaders. Okay. And we're going to multiply it by the malaria T. That we just solved four. So when we set it up this way, the units cancel for leaders and we're left with moles. Okay. And so once we solve for this, we're going to get 0. 04, three four moles of ions. Okay, So now that we know how many moles of ions we have, we're going to determine the mass of chlorine and sodium chloride. Okay, So we have A 1. g mixture of sodium chloride, sodium nitrate and magnesium nitrate. And we're told it is 12.41%. Or if we convert this to a decimal is going to be . chlorine by mass. Okay, so our mass for chlorine Is going to equal. So we have 1. g and we're going to multiply this by 0.1241. And when we do that this gives us 0.130305g of chlorine. So now we can solve for the mass of sodium chloride. Okay, so for a mass of sodium chloride we're going to use the grams of chlorine. We just solve for which is 0. g of chlorine. And now we're going to go from grams of chlorine to moles of chlorine using molar mass and one mole of chlorine. There is 35. g of chlorine. Okay, So our units for grams of chlorine cancel. And we're left with moles of chlorine. So now we're going to go from molds of chlorine, two moles of sodium chloride by using the multiple ratio. Okay, so the multiple ratio is in one mole, in one mole of sodium chloride, there is one mole of chlorine. So now our units four moles of chlorine cancel. And we're left with moles of sodium chloride. Now we're going to go from moles of sodium chloride, two g of sodium chloride Using the molar mass of sodium chloride. In one mole of sodium chloride, There is 58.44 g of sodium chloride. Our units for moles of sodium chloride cancel. And now we're left with grams of sodium chloride, which is what we're looking for. We're looking for the mass of sodium chloride. So let's go ahead and do this calculation for massive sodium chloride. When we do this calculation, what we're going to get is zero 214, grams of sodium chloride. Now that we know how many grams of sodium chloride we have. We're going to determine the number of mole ions from sodium chloride, sodium nitrate and magnesium nitrate are moles of ions from sodium chloride will start there. Okay, so more ions from sodium chloride. So our starting point is going to be the g we just calculated. So 0. g of sodium chloride. Alright, so now we want to go from grams of sodium chloride, two moles of sodium chloride. So in We're going to need the molar mass here and one mole of sodium chloride. There is 58. g of sodium chloride. Okay. And now we're going to go from moles of sodium chloride to moles of ions. Okay, so in one mole of sodium chloride, how many more ions do we have? We have two more ions. Okay, so when we do this, we'll look at our units to make sure they cancel grams of sodium chloride, cancel moles of sodium chloride cancel. So now we have more ions and that's what we're looking for here. So once we do this calculation, what we're going to get for the more ions Is 0.00735. one more ions. Alright, so now are more ions from sodium nitrate. And magnesium nitrate is going to equal the difference between are starting more ions. And our So remember at the top we got the total number of moles from the malaria T. So our total number of moles more ions minus the more ions that we just calculated. We're going to take the difference between those two. Okay, so we just found the more ions from sodium chloride. So now we'll say the more ions from the sodium nitrate for sodium nitrate and magnesium nitrate equals. So our total more ions, which is this value here, we're gonna start there. So we have 0. more ions - What we just calculated for the sodium chloride which is 0. 0735. one more ions. Okay, this is going to give us 0. mole ions. So let's let we're going to let X equal the mass of sodium nitrate. And we're going to let y equal the mass of magnesium nitrate. So we know that we we know what the total mixture weight is. And so we're going to say X plus Y is going to equal the total mixture weight. And we're told in the problem that is 1. g for the mixture. Okay. And we're going to subtract The g of sodium chloride. So we're going to subtract 0. g of sodium chloride. So this is going to give us 0. g of sodium nitrate and magnesium nitrate. Okay, so if we isolate or try to solve for one of our variables, let's try to solve for X. What this is going to become is X is going to equal zero point minus Y. Okay, so I'm going to do some plugging in. So for the left side of our equation we're going to get X over 85 times two ions plus. Why? over .63 times three ions equals 0. moles. Okay, and that is us multiplying by the molar masses. Alright, so, having it set up that way, we're going to solve for y. Okay, so, solving for Y Equals 0. 02237 g of magnesium nitrate. And solving four x equals 0. -6. Okay, so we can solve for X. X. is going to equal 0. g zero 702237 g equals 0. g of sodium nitrate. So now we can calculate the mass percent. Okay, so the mass percent of sodium chloride is going to equal The g of sodium chloride which is 0. 810 g of sodium chloride. And this is going to be over the mass of the sample. And the mass of the sample is 1.05 g. And this is going to be multiplied by 100%. So, what we're going to get for the mass percent of sodium chloride is .46%. So, we're going to go ahead and do that for sodium nitrate and magnesium nitrate. So, for sodium nitrate we'll write the mass percent. For sodium nitrate is going to equal The g of sodium nitrate which is 0. g of sodium nitrate. And this is going to be over the mass of the sample, which is 1. grams Sample Times 100%. And this gives us 12.66 as the mass percent for sodium nitrate. And lastly, we need the mass percent for magnesium nitrate. Okay, so the mask for the magnesium nitrate is 0. 02237 grams of magnesium nitrate. And this is over the mass of the sample. So the mass of the sample is 1.05 g. Okay, And this is multiplied by 100%. So the mass percent for magnesium nitrate is 66.88%. So let's go ahead and highlight our final answers. The mass percent for sodium chloride is 20.46%. The mass percent for sodium nitrate is 12.66% and the mass percent for magnesium nitrate Is 66.88%. That is it for this problem. I hope this was helpful.

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