Calculations with the unit cell of $NaCl$

A monocristal of NaCl has a mass of 410 g; Immersed in a beaker filled to the brim with petrol, it causes overflow of 190 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.

$M_{NaCl}$ $=$ $35.3$ $+$ $23$ $=$ $58.5\frac{g}{mol}$ $\rho_{NaCl}$ $=$ $\frac{m}{v}$ $=$ $\frac{410}{190}\frac{g}{cm^3}$ $V_{molar}$ $=$ $\frac{M}{\rho}$ $=$ $\frac{58.5\cdot 19}{41}cm^3$ There are $6\cdot 10^{23}$ ions $Na^+$ per mole and 4 ions $Na^+$ in one lattice cell. So we have $\frac{6\cdot 10^{23}}{4}$ lattice cells per mole $NaCl$ $V(maille)$ = (volume of one mole)/( number of cells in one mole) = $\frac{58.5\cdot 19\cdot 4}{41\cdot 6\cdot 10^{23}}cm^3$ The cell is so a cube with volume: $V(maille)=a^3$.
Therefore: $a$ $=$ $V(cell)^{\frac{1}{3}}$ $=$ $(\frac{58.5\cdot 19\cdot 4}{41\cdot 6\cdot 10^{23}})^{\frac{1}{3}}$ $=$ $5.65\cdot 10^{-8}cm$