# CsBr crystallises in a body-centred cubic (bcc) lattice. The unit cell lengt is 436.6pm. Given that the atomic mass of Cs= 133u and that of Br=80u. What will be the density of CsBr? a) 0.425g/cm^3 b) 8.25g/cm^3 c) 42.5g/cm^3 d) 4.25g/cm^3

d)4.25g/cm3

Using the atomic masses of Cs and Br, determine the mass of one atom of CsBr.

133u plus 80u equals 213u.

213 g/mol

convert pm to cm first.

Compute the unit cell's volume.

You now have to be aware of

mass/volume equals density.

Since we already know the mass, please convert amu to grams.

35369e-22 = 213amu

3 sig figs = 4.25g/cm

Option D is therefore accurate.

An additional method is

where volume is equal to a^3 = atomic radius^3.

Z is the number of molecules per bbc. The bbc is usually 2 because bbc's are of atoms and elements have 1 atom which makes up a molecule. However, for this bbc, Z = 1 because 1 bbc has 2atoms, which here means 1atom of Cs and 1atom of Br giving a total of 1molecule.

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The density of CsBr is approximately 4.25g/cm^3 (Option d).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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