A cube-shaped crystal of an alkali metal, 1.62 mm on an edge,
wa...

Question

A cube-shaped crystal of an alkali metal, 1.62 mm on an edge, 
was vaporized in a 500.0 mL evacuated flask. The resulting 
vapor pressure was 12.5 mm Hg at 802 °C. The structure of 
the solid metal is known to be body-centered cubic. 
(c) What are the densities of the solid and the vapor in 
g>cm3?

Accepted Answer

Hello. In this problem, we are told in a 350 million evacuated flask, a cube shaped crystal of an Alkali metal with an edge length of .936 mm was vaporized. The vapor pressure that resulted was 13.7 millimeters of mercury at 789 degrees Celsius. It is well known that the metal has a body centered cubic structure. Were asked to determine the density in terms of grams per cubic centimeter of both the solid and the vapor. Let's begin by going through the steps that we will take to solve this problem. So we have a alkaline metal that's been vaporized and make use of the ideal gas equation. So we'll use the ideal gas equation to find our molds of metal and so will represent the metal by a capital M. We can then make use of our molds of metal and the volume of the crystal to find the volume of the unit cell. We can then find the edge length of the unit cell, from the volume of the unit cell and there we can determine the radius of an atom and then the identity of our alcohol metal and then we'll use the molar mass of the metal, which we will now know from Step three to determine the density of the solid. And then we will use the roles of the metal and the volume of the flask to determine the density of the vapor. Let's begin making use of the values were given by converting our volume and temperature to the desired units. So we're gonna take the volume of our flask and we're going to convert it from milliliters leaders, so 1000 mL is equal to one liter. So set ups are units and milliliters cancels. This works out to 0.350 liters. And then the temperature, since we're going to be dealing with the gas, we want the temperature in kelvin. So we have 7 89 degrees Celsius plus 273 is 1062. Kelvin will now make use of the ideal gas equation. Find the molds of our metal metal will then will be equal to the pressure times volume divided by the gas constant, divided by temperature, are pressure is given in the problem statement as 13.7 millimeters of mercury are volume is 0.350 leaders. So the gas will expand to completely fill the flask and temperature 1062 kelvin make use of our value of 0.8206 leaders. Atmospheres per kelvin mole. And then we're going to convert our units of pressure from millimeters of mercury, two atmospheres, there's 760 millimeters of mercury and everyone atmosphere. So we're going to ensure that our units cancel millimeters of mercury cancels, atmospheres cancels kelvin cancels and leaders cancels. So we're left with units of moles or moles. Works out to 7.240 have seven minus five moles of our metal are crystal is a cube So we can find its volume by taking its edge length and cubing it. We're told the edge length of the crystal was 0.936 millimeters cube. That we're going to convert our volume two cubic centimeters. So we have 1000 millimeters is equal to one m. We're gonna cue both of those so that our units cancel and one m is equal to 100 centimeters. We're gonna cube both of those as well. So now our units of meters cancels. So the volume of our crystal works out to 8.200 times 10 to the minus four cubic centimeters from that information. And recall that for a body centered cubic unit cell, we have two atoms, one in the center and the eight corners make up another one. So using the information that we calculate above, we can find the volume of our unit sale, take the volume of our crystal and divided by moles of our metal and we'll make use then of avocados number. So one mole of metal contains six point oh times 10 to the 23rd metal atoms. And we set it up so that our moles of metal cancels. And we have two metal atoms in every one BCC unit cell and our atoms cancels. So we're left with volume per unit cell. This works out to 3.76 times 10 to the negative cubic centimeters per unit cell. Moving on to the next step, we're going to now find the edge length so we can find the edge length of our unit cell. If we find the cube root of the volume of our unit cell. So taking the cube root of the volume of our unit cell, which is 3.76 times 10 to the minus 23 cubic centimeters. We find that the edge length of our unit cell is 3. times 10 to the -8 cm. Recall that for a body centered cubic unit cell, the radius of an atom is equal to the square root of three, divided by four times the edge length. So we have the square to three divided by four times the average length, which was 3.35 times 10 to the -8 - centimeters. And this works out to then 1.45 times 10 to the minus eight centimeters will convert our radius from centimeters to PICO meters. Since this is generally how the radius is reported in tables. So 100 centimeters it's one m. Make sure our units centimeters cancel. And then one m is equal to one times 10 to the m. And are units of meters cancels. So our radius works out to 145 PICO meters. If you refer to a table of lithium, sorry, alkaline metal Adams, you'll find that lithium has a radius of 145 m. The next step, then we're gonna make use of the molar mass of lithium lithium has a molar mass of 6.941 g per mole. We have two atoms Of lithium and every body centered cubic unit cell, one mole of lithium containing 6.02, 2 times 10 to the 23rd lithium atoms. So we set up that conversion factor. So our units lithium atoms cancels and making use of the molar mass of lithium. One mole of lithium has a mass of 6. g of lithium lithium cancel. So our massive lithium is 2.305 times 10 to minus 23 g of lithium. The density event of our solid. We take the massive r lithium atoms and we divide by the volume of our unit cell, which we found to be 3.76 times 10 to minus cubic centimeters. This works out then to a density of 0.613 grams per cubic centimeter. This is then the density of the solid. And moving on now we'll find the density of the vapor. We'll make use of the molds of our metal that we found. We now know that that is lithium. So we have 7.240 times 10 to the minus five moles of our metal which we again now know is lithium divided by volume of the flask since the gas will expand to completely fill the volume of the flask, That's the volume of our gas and make use of the Molar Mass. one mole of lithium Has a massive 6.941 g of lithium. So are, you know, Mosul, lithium cancel the name to the proper units of volume. One liter is equal to 1000 mL. So our leaders cancels and one millimeter is equivalent to a cubic centimeter. So units milliliters cancels. This works out to them 1.4, 4 times 10 to the - grams her cubic centimeter. This is the density of the vapor. So we were able to determine the identity of our alkali metal, knowing that we can then make use of its molar mass as well as a few other things to determine the density of the solid and then the density of our vapor. So the density of our alkali metal solid, which is lithium is 0.613 g per cubic centimeters. And the density of our vapor is 1.44 times 10 to the minus six g per week centimeters. Thanks for watching. Hope this helps

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