Sodium has a density of 0.971 gcm3 and crystallizes with a
body...

Question

Sodium has a density of 0.971 g>cm3 and crystallizes with a 
body-centered cubic unit cell. What is the radius of a sodium 
atom, and what is the edge length of the cell in picometers?

Accepted Answer

Hello everyone today, we have the following problem, potassium crystallizes into a body centered cubic unit cell with the density of 0.862 g per centimeter cubed, determine the edge length of the cell and radius of the atom in PICO meters. So the very first thing that we must do is we must find our mass of our potassium atom. And to do so we have to start with the molar mass of the of potassium. So according to the periodic table, the molar mass of potassium is 39. g per mole. And we must multiply this by the conversion factor, That one mole is equal to 6.02 times 10 to the 23rd Adams. In doing so, our units were almost canceled out and we were left with 6. 926 times 10 to the negative 23rd grams per Adams. Now it was stated that the potassium isn't a body centered cubic unit cell. And so since our potassium is in a body centered cubic unit cell, We can say that we can fit two potassium Adams in a unit cell knowing that we can go ahead and determine our mass directly of our unit cell. And so our unit cell mass is going to be equal to The two potassium atoms times the conversion factor that we just sold for the 6.4926 times 10 to the negative 23rd grams per atoms. And our units for Adams can cancel out to where we are left with 1.2985 times 10 to the negative 22nd grams. So we're gonna hold on to that number there. Next we need to find the unit cell volume. So we've solved for its mass and now we must solve for its volume. And the unit cell volume is going to be The mass of the unit cell over its density. And luckily we have both of those values. So the mass we just calculated for was 1. 85 times 10 to the negative 23rd grams. And our density given to us in the question is 100.862 grams per centimeters cubed. Our units for grams will cancel out we will be left with 1.5064 times 10 to the negative 22nd cubic centimeters. And so we'll hold on to that number as well with these values that we just calculated for. We can actually solve for our unit cell edge. So we're gonna solve for our unit cell edge length and then it's going to be in terms of D as our unit as our variable. And so that's going to be three radical with our volume which was 1. 064 times 10 to the negative 22nd cm cubed. When we solve for this we actually have to take into account that we have to convert this to PICO meters. So we're going to get 5.3 times 10 to the negative eight centimeters. But we have to convert to PICO meters. And so to do that, we're gonna use the conversion factor that one centimeter here. The denominator is equal to one times 10 to the negative two m. And that one PICO meter in the numerator because that's what we want to find is equal to one times 10 to the negative m. Units for meters and centimeters cancel out. And we're left with 532 . m. And so that's going to be our edge length in pickle meters here. And then lastly we have to find the radius of the atom and the radius is going to be equal to our four R or four times the radius is equal to the square root Of three times our edge length. And so to solve for this, We're simply going to take radical to the 4th for both of these values and end up with a radius being equal to Three times our edge length that we found, which was 532. divided by four. And so we're going to end up with a value of 230. m as our radius of our atom Overall, I hope that it's helped And until next time

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