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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Hello. In this problem, we are told a 1.127 g chunk of an unknown metal was dissolved in excess Acquis hydrochloric acid, the volume of evolved hydrogen gas was major to be 720 mL at 23 degrees Celsius and 230.954 atmospheres. The reaction mixture was heated to drive off the excess hydrochloric acid and water. Colorless, crystalline solid metal fluoride was obtained. The solid was then re dissolved in 45 g of water to make a clear solution. The freezing point of the solution was measured as -3.50°C. We're being asked to determine the molar mass and the molecular formula for our metal chloride. So first step then let's begin by writing our reaction and finding the moles of hydrogen gas that was produced using the ideal gas equation. So we have our metal which is a solid, react with hydrochloric acid, give us Arjun gas and or middle fluoride. Making use of the ideal gas equation. Then our moles of hydrogen is equal to the pressure, times the volume divided by the gas constant divided by temperature. Our pressure were given as 0. atmospheres are volume was ml Convert that to leaders. So 1000 ml Is equal to one L. Our units and ml cancels. This works out to 0.720 L. Our temperature is 23. The C will add to 73.15 And this works out to them to our 96.15 Kelvin. So plugging these things into our equation for moles we have our pressure volume provided by our temperature And then our gas constant of 0.08-06 L, atmospheres per Kelvin mole. So checking that our units cancel, so kelvin cancels, leaders cancels and atmospheres cancels and we're left with moles. This works out to then 0.0-8- rules of hydrogen so we'll shrink this up quick so that we can have more room for the rest of our problem. And so step two, then we're now going to make use of are moles of hydrogen defined a massive Maureen and our mass of our metal chloride. So we have zero point 0-8-6 moles of hydrogen gas Based on the balanced reaction shown in step one they have x moles Hydrofluoric acid react for every X over two moles of hydrogen. One more. Huh? Hydrofluoric acid has one mole of chlorine and one mole of chlorine Has a mass of 35.453 g. So checking that our units cancel our moles of hydrogen, cancels moles of hydrochloric acid cancels and moles of foreign cancels. So this works out to then 2. grams of boring. So we'll take a minute to shrink this one up as well. And so now we're going to calculate our total mass. So we have 2.004g of chlorine And we were told in the problem that we had 1.127 g of our metal. So our total mass then works out to 3.131g. So this is a total mass of our metal cord. Our next step then we're gonna use the freezing point depression formula to find or oz morality. So I times end M and find the moles of our metal chloride and solution. So the change in the freezing point then will be equal to the van Hoff factor, times the freezing point depression constant, times morality. The freezing point depression constant for water then is equal to 1.86 degree Celsius per allow our change in temperature freezing then is equal to the freezing temperature for pure water minus the freezing temperature for the solution Which is 0.00°C -2.50°C. So to change the temperature Is equal to 3.50 three Celsius. So solving now for osmolarity, it's equal to the change in our freezing point divided by the freezing point depression constant. This works out to 1. wow, Which is 1.88 Nice of our metal chloride in solution invited by programs of wir shrink this up a bit. Next. Then we are told that we dissolve the metal chloride in g of water. So we'll make use of our 45 g of water. And and the as morality to determine our moles of metal chloride in solution. So 1000 g is equal to kilograms. So we're going to write 1000 g of the denominator. You see then that our grams of water cancels. And we're left with then 0.08 or six roles of our middle board in solution. So keep in mind that this is equivalent to I times our moles. Then in the next step we are going to find the step four, we're gonna find the molecular formula And the molar mass. So we have from step two that are massive flooring was 2.004g. And we'll convert that to moles. So one mole Chlorine has a mass of 35 453 grams of chlorine grams of chlorine, cancel. We're left with multi chlorine. This works out to 0. bottles of chlorine. We're now going to determine how many moles metal we have. So we'll take our M.olds of our metal chloride in solution. So 0.0846 and subtract off our mold deploring. This works out to 0.0 2807. Most of our metal. You can now find the like the formula will take our moles of boring zero point 05653 divided by our moles metal 0. 2807. And this works out to two. So our molecular formula then is m. c. l. two. So that was one thing we were asked to determine was the formula for our metal chloride. We'll shrink up these last few things a bit. They have a little more space. And then our last thing we're going to do is then determine our molar mass. So returning to our reaction in step one We see that one mole of middle then reacts to form one mole of our metal chloride. And so from step four we found our molds metal to be equal to 0.028 07, which is equal to then our roles of our middle chloride. And we found that the mass of our metal chloride Was 3.131 g. We'll divide that by our molds and this works out to g per mole. So this is our solar mass for a metal chloride. So we found that our metal chloride has a formula of M. C. L. Two and its molar mass is 111 g per mole. Thanks for watching. Hope this helps

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Key Concepts

Ideal Gas Law

Colligative Properties

Stoichiometry