Calculations with the unit cell of cesium chloride

A monocristal of CsCl weighs 260 g. Immersed in a beaker filled to the brim with petrol, it causes overflow of 65.5 cm³ liquid. Calculate (answers at the bottom):   - the molar mass of CsCl.   - the mass density of CsCl.   - the molar volume of CsCl.   - the number of cations (or anions) per unit cell.   - the number of unit cells per mole.   - the volume $V$ of one unit cell.   - the length of the edge $a$ of one unit cell.

$M_{CsCl}$ $=$ $35.3+132.93=168.4\frac{g}{mol}$ $\rho_{CsCl}=\frac{m}{v}$ $=$ $\frac{260}{65.5}\frac{g}{cm^3}$ $V_{mole}=\frac{M}{\rho}$ $=$ $\frac{168.4\cdot 65.5}{260}cm^3$ There are $6\cdot 10^{23}$ ions $Cs^+$ per mole and 1 ion $Cs^+$ per unit cell. So there are $\frac{6\cdot 10^{23}}{1}$ unit cells per mole $CsCl$ $V(unit cell)$ = volume of one mole/number of unit cells per mole = $\frac{168.4\cdot 65.5\cdot 1}{260\cdot 6\cdot 10^{23}}cm^3$ The unit cell is cubic with a volume of $V(unit cell)$ $=$ $a^3$. Therefore: $a$ $=$ $V(unitcell)^{\frac{1}{3}}$ $=$ $(\frac{168.4\cdot 65.5\cdot 1}{260\cdot 6\cdot 10^{23}})^{\frac{1}{3}}$ $=$ $4.13\cdot 10^{-8}cm$