[Ann1234]

Hello all! I have this problem that I am not sure on how to solve- I checked online and I found that the answer could be 5.7x0.93^3 but I don't have an idea on why this is correct and also I would like an explanation on how to do this problem step by step- thanks in advance!

At low temperatures, xenon freezes to a solid having a density of 5.7 g/cm3. Below 84 K, argon also freezes to a solid that has a similar structure but different volume and mass. If the volume of each unit cell in argon is 93 percent of the xenon unit cell, what is the predicted density of solid argon?

[UG]

Solid xenon packs as a face centered cubic (according to my periodic table), with this information, you can work out the mass of one unit cell of xenon, have you learnt how to do this? Because you know the density, you can then work out the volume of each xenon unit cell. Once you find the volume of the xenon unit cell, you can easily find the volume of the argon unit cell. Argon also packs as a face centered cubic solid so you can calculate the mass of one unit cell of argon. Finally the density can be found. See if this helps.

[Ann1234]

well I got something-

1- with the density of Xe, I isolated cell volume using this formula

D = ZXM/a^3xN where a^3 = cell volume, N= avogadro number, M = molar mass, Z = 4 for FCC

2- Then I calculated the 93% of Xe cell volume to get Ar cell volume which is 1.53E-22

3-With the value of Ar cell volume, the Ar molar mass, Z, and avogadro # I calculated Ar density using the formula D = ZXM/a^3xN

I got 1.87 g/cm^3 as answer - not sure if is correct but at least makes sense to me. Thank you!

[UG]

Yea that looks ok. Though I think 1.53E-22 is the volume of the xenon unit cell, probably just a typo as your answer is similar to mine.

[Ann1234]

oh yes, 1.53E-22 is Xe cell volume, you are right. And I got 1.42E-22 as Ar cell volume.