Platinum crystallizes with the face-centered cubic unit cell. The radius of a platinum atom is 139 pm. Calculate the edge length of the unit cell and the density of platinum in g/cm3 .

Video transcript

Welcome back everyone to another video plats crystallizes with the face centered cubic unit cell. The radius of platinum atom is 139 Peters calculate the edge l of the unit cell and the density of platinum in grams per cubic centimeters. We're given four answer choices. A 36.9 B 21.3 C 7.83 ND 4.69. So we wish to begin this problem by simply visualizing the structure of the FCC or face center cubic cell. We are just looking at the front because it tells us a lot. We have our half of an atom in front and we have an eighth of an atom at each corner. And now what we notice is that we can actually introduce a relationship between the radius and the edge land. So let's suppose that the edge land is a right, we have a square and let's suppose that we have a radius, we can essentially extend it along a diagonal of that square. So how many radiate do we have? Well, we have 1234 and we can apply the Pythagoras theorem saying that if we take or R and if we square it, that's our hypotenuse of a right triangle, applying the Pythagorean theorem, this should be equal to A squared plus A squared. And therefore 16 R squared is equal to two A squared. And eventually, if we are solving for the edge land, we get a squared equals eight R squared. So A since it's about the value is just two squared of two R, right, because square of beta two screwed up two. So we have our first relationship that's really important for us, we can get our edge land. We are going to take two screwed up two and multiply that by 139 petters. So our edge land in this case will be 393.15 Peters. Now, once you valid the density and we know that density is equal to mass divided by volume. Now we can take one cube and we have to understand that we have a total of four atoms in one cell, right? We have eight eighths of an atom at each corner of that cube. So that gives us one and then we have six halves. So essentially this gives us a total of four atoms in one cell. What we're going to do is take four multiplied by the molar mass of platinum and divide by the avocados number. Now, why are we doing this? Well, essentially, if we take the molar mass of platinum and divided by the avocados number, we get the mass of one atom and we're multiplying by four to get the total mass of four atoms in one cubic cell. And now we have to divide it by a volume. Since we have a cube, the volume of such a cube would just be a cubed. And we have our final formula, the density would be four multiplied by the molar mass of platinum and divided by the product between a cubed and the avocados number. So let's plug the variables or so we have four. Now, the question is, what's the molar mass of platinum? Well, we need to use the periodic table 195.09 g per mole. Now, on the bottom, we have to think about the unit conversion. So if we went to cube A which is 393.15 Peters, we first of all want to convert it into centimeters. So we're taking Peters multiplying by our conversion factor. First of all, let's go into meters. 1 m is equivalent to 10 to the power of 12 Peters and then we need centimeters. So 1 m is a 100 centimeters. We're going to CD A. And then we also need to include the avocados number 6.022 multiplied by its sense to the power of 23rd mold to the power of negative first. Now let's validate the result here when we do the math. We end up with 21.3 g per centimeter cube, which essentially corresponds to the answer choice B in the multiple choice. That'll be our final answer. And thank you for watching.