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. (b) Use the data in Figure 5.19 to identify the alkali metal.

Video transcript

Hello. In this problem, we are told in a 350 million to evacuate flask, a cube shaped crystal of an alkali metal with an edge length of 3500. millimeters was vaporized. The vapor pressure that resulted was 13.7 millimeters. Mercury at 789 degrees Celsius. It is well known that the metal has a body centered cubic structure. Were asked to identify the metal. Let's begin by outlining the steps that will take to arrive at the identity of the middle. First thing we'll do since we vaporizes the metal, we can make use of the ideal gas equation from the ideal gas equation, we will identify the moles of our alkali metal which will refer to as just capital M moving forward. We can then make use of the molds of our metal and the volume of the crystal to determine the volume of the unit cell. And then in step three we can find the edge length then of the unit cell. From there, the radius of an atom and finally the metal identity. Let's begin by doing some conversions with the values that were given above. So the volume of our flask we're told is mL. So we'll convert that volume two leaders. It's 1000 ml Is equal to one leader. We set up our conversion factors. So ml canceled this works out to 0.350 l And then the temperature that we're given is in C will convert that to Kelvin. Since we need Kelvin, we're making use of the ideal gas equations, we have 789°C will add to 73 to convert it to kelvin and that works out to 1,062 Kelvin. Now making use of the ideal gas equation molds of the metal then is equal to the pressure times volume divided by the gas constant, divided by temperature. Our pressure, we were told was 13.7 millimeters of mercury. Our volume was .350 L. Since the gas will expand to completely fill the volume of the flask, We have our temperature which is 1,062 Kelvin. We have in our value 0.08-06 leaders, atmospheres for Kelvin Mole and one atmosphere Is equal to 760 of mercury. So we need to ensure that our units are canceling, so mm of mercury cancels atmospheres cancels, kelvin cancels and leaders cancels. So we're left with units of moles, which is what we're after. So the moles of our metal works out to 7.240 times minus five. Now we'll calculate the volume of our crystal volume of our crystal Is equal to the edge length cubed. We're told that the edge length of our crystal was .936 mm cubed. And we're gonna to convert this to cubic cm. So 1000 Is equivalent to a meter. So we're gonna cube both of those so that our units will cancel and one m is equal to 100 cm. We're going to cue both of those and are units of meters cancels. The bottom of our crystal works out to 8.200 times 10 to the minus four cubic centimeters. Now we can find the volume of our unit cell. This will be equal to them, our volume of our crystal provided by the molds of our metal and making use of avocados. Number one mole of our metal contains six point oh 22 times 10 to the 23rd metal atoms we set up so that moles of our metal cancel. And within a body center cubic unit cell. Recall that one unit cell contains two metal atoms. So again this is for a body centered cubic unit cell. And now our units of Adams cancels and we're left with cubic centimeters per unit cell. This works out to 3.76 times 10 to the minus 23 cubic centimeters per unit cell. You can find the edge length of our unit cell taking the cube root of our volume. That would be equal to then the cube root of 3.76 times 10 to the minus 23 centimeters cubed. And so our edge length works out to 3. times 10 to the -8 cm. We called it for a bodied centered cubic unit cell. That the radius then of an atom is equal to the square root of three, divided by four times the edge length. So we have then the square to three divided by four times our edge length. With three deaths. Found to be 3.35 times 10 to minus eight centimeters. This works out to 1.45 times 10 to the -8 cm. We're gonna take our radius then and convert it from units of centimeters, two people meters, which is provided in the problem statement And one m is equal to 100 cm. Our units of centimeters cancels And one m is equal to one times 10 to the 12 km. So we said that these are units of meters cancels. This works out to then Radius of 145 m. If we look up at the top then for the identity of our Alkali metal, we find then that the element with a radius of km is lithium. So lithium is then the correct answer. Thanks for watching. Hope this helped