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Ch.13 - Solutions & Their Properties

Chapter 13, Problem 149f

Treatment of 1.385 g of an unknown metal M with an excess of aqueous HCl evolved a gas that was found to have a volume of 382.6 mL at 20.0 °C and 755 mm Hg pressure. Heating the reaction mixture to evaporate the water and remaining HCl then gave a white crystalline compound, MClx. After dis- solving the compound in 25.0 g of water, the melting point of the resulting solution was - 3.53 °C. (f) What is the identity of the metal M?

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Hello everyone today. We have the following problem. A 1.127 g chunk of an unknown metal was dissolved in excess. Acquis hydrochloric acid. The volume of the evolved hydrogen gas was measured to be 720 mL at 23 degrees Celsius and 230.954 atmospheres pressure. When the reaction mixture was heated to drive off excess hydrochloric acid in water. A colorless crystalline solid was obtained. The solid was then re dissolved in 45 degrees or 45 g of water to make a clear solution. The freezing point of the solution was measured at negative 3. degrees Celsius, identify which metal it is. So the very first thing we need to do is we need to find the number of moles of our hydrogen gas. And so what we're gonna do is we're gonna use a derived version of the ideal gas law. And so that's going to be that the number of moles is equal to the pressure times volume, divided by the gas constant, times our temperature. So our pressure was 0.954 atmospheres. Our volume was 720 million years. However, We're going to need to convert that into leaders. So we're just gonna multiply by the conversion factor that we have one militia leader per every 10 to the negative third leaders so that our middle leaders will cancel out. And we're going to divide that and we're going to divide that first r gas constant, which is 0. Leaders times atmospheres per mole times kelvin. And then we're gonna multiply by the temperature which is 296.15° Kelvin. And we got that temperature because we added 273 to our 23°C. And when we did that we got the number of moles was equal to 0. moles of our hydrogen. And so we're gonna hang on to that number there. So we need to write a hypothetical reaction next that best represents the number of moles. And then we need to find the mass of chlorine that reacted with this hypothetical m so how can we do that? Well, our hypothetical equation is going to have to look something like this. We're going to have to have em we're going to have to add that to x number of moles for our hydrochloric acid and that is going to give us some number over to for our moles of our hydrogen gas and then the colorless crystalline solid here. So to find the moles of our chlorine Or chloride, we have to take the number of moles of hydrogen gas that we got initially. So that's going to be 0.028 most of our hydrogen gas. And then we're gonna multiply that by our multiple ratio here so that we have half a mole of our hydrogen gas per one mole of our hydrochloric acid. That's our multiple ratio. And we use the coefficients for that. We then multiplied by the ratio that we have one mole of hcl and in hcl we only have one chlorine. So we're going to say that we have one mole of chlorine when our units cancel, we're going to have 0. moles of chlorine. Next we're going to use that to find the mass of our chlorine. So we're gonna take our 0.0565 moles of chlorine And multiply that by the molar mass of Chlorine, which on the periodic table was one mole of chlorine is equal to 35. g of chlorine. And this is going to give us two g of chlorine. So that's our mass. And that's the number of moles that we have. Third, we are going to use that mass of M and chlorine to find the total mass of the product that we have. So the mass of our crystalline structure, it's going to be equal to our unknown metal chunk mass which was 1.127 g plus the mass of chlorine we just sold for which was to And that is going to give us 3.13 g of our crystalline solid, which is just the M C. L. X. Next we are going to use the freezing point depression formula to find the total morality of the solution. And then the total number of moles of our chris, Lynn's solution or crystalline structure in the solution. So we have the formula that the freezing point depression is equal to our negative vant Hoff factor times our freezing point constant times our morality or m Since the solvent is water, the value can be looked up in a reference table. And so our KFR freezing point constant here is going to be 1.86°C times kilogram per mole. And so the freezing point of pure water is zero. So to solve for our freezing point or freezing point, we're going to take the negative 3.50°C. That was in the question stem and subtract that by the freezing point of pure water giving us negative 3.50 degrees Celsius as our change in our temperature. So now we just need to plug in these values, we're gonna have negative 3.50 degrees Celsius is equal to negative I. M. Times R K. F value, which is 1.86. That is gonna give us that are I our I times R. M. Or event. Half factor times our morality is going to be equal to 1.88 moles per kilogram. And so we're gonna save that number for later. What we can do now now that we have the total morality of the ions and we have the total mass of the solvent. We can now find the total moles of our crystalline solid in the solution. So what we're gonna do is we're going to take our mass our 1. moles per kilogram and multiply that by the conversion factor that we have our molds of our crystalline structure per one kg of water. We're then gonna multiplied by the factor that one kg of water is equal to 10 to the third grams of water. And then lastly we're gonna multiply by our massive water that we have which is g when our units cancel out, we were left with 0. moles total of our crystalline solid. Now using the total moles that we found and chlorine, we're going to find the moles of em. And so the moles of em can be found by taking the moles of our crystalline structure and subtracting them by the moles of our chlorine. We're gonna take our 0.085 moles. It's attracted by R .0565 moles of the chlorine. We found earlier to get 0. moles of em. And so using our total moles for chlorine animals of em, we can find the molecular formula So we can do that by finding the mole ratio, find the bull ratio between our chlorine and our m So we're gonna take the moles of our chlorine which was 0.0565 moles. And our mole's of em. And divide that by that. So 0.028 moles. That's going to give us approx. two. That's our mole ratio. And so the formula for the compound is M. C. L. Two. This is our formula here using this mass that we have and the moles, we can find the molar mass. So to find the molar mass of R. M. C. L. Two. We're going to take our mass which was 3. 36043 g. We're gonna divide that by the M.olds that we have which is 0.028 moles. This is gonna give us our molar mass of 111.22 g per mole. What we can do next is we can subtract the molar mass of chlorine in the compound from the molar mass of the compound to find the molar mass of M. So we have our molar mass of em being equal to 1 11.22 g per mole subtracted by The two moles that we have of chlorine times the molar mass of chlorine itself which was 35.45 g per mole. And this is gonna give us 40. g per mole. And from this we can identify the metal. So the formula that we said was M. C. L. Two. So the metal must be die violent meaning it must be in our group two on the periodic table. And so this molar mass is going to directly resemble that of calcium on the periodic table, so that is what our metal is. And with that we have answered the question overall, I hope it has helped, and until next time.
Related Practice
Textbook Question
Addition of 50.00 mL of 2.238 m H2SO4 1solution density = 1.1243 g>mL2 to 50.00 mL of 2.238 M BaCl2 gives a white precipitate. (b) If you filter the mixture and add more H2SO4 solution to the filtrate, would you obtain more precipitate? Explain.
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Textbook Question

Treatment of 1.385 g of an unknown metal M with an excess of aqueous HCl evolved a gas that was found to have a volume of 382.6 mL at 20.0 °C and 755 mm Hg pressure. Heating the reaction mixture to evaporate the water and remaining HCl then gave a white crystalline compound, MClx. After dis- solving the compound in 25.0 g of water, the melting point of the resulting solution was - 3.53 °C. (c) What is the molality of particles (ions) in the solution of MClx?

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Textbook Question

Treatment of 1.385 g of an unknown metal M with an excess of aqueous HCl evolved a gas that was found to have a volume of 382.6 mL at 20.0 °C and 755 mm Hg pressure. Heating the reaction mixture to evaporate the water and remaining HCl then gave a white crystalline compound, MClx. After dis- solving the compound in 25.0 g of water, the melting point of the resulting solution was - 3.53 °C. (e) What are the formula and molecular weight of MClx?

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Textbook Question

A compound that contains only C and H was burned in excess O2 to give CO2 and H2O. When 0.270 g of the com- pound was burned, the amount of CO2 formed reacted completely with 20.0 mL of 2.00 M NaOH solution according to the equation 2 OH-1aq2 + CO21g2 S CO 2- 1aq2 + H2O1l2 When 0.270 g of the compound was dissolved in 50.0 g of camphor, the resulting solution had a freezing point of 177.9 °C. [#Pure camphor freezes at 179.8 °C and has Kf = 37.7 1°C kg2>mol.] (a) What is the empirical formula of the compound?

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Textbook Question
A solution prepared by dissolving 100.0 g of a mixture of sugar 1C12H22O112 and table salt (NaCl) in 500.0 g of water has a freezing point of - 2.25 °C. What is the mass of each individual solute? Assume that NaCl is completely dissociated.
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