How many units cells are there in 1.00g cube shpaed ideal crystal of AB `(M_(W) = 60) which has a NaCl type lattice?

Options

A. `6.02xx10^(23)`
B. `2.50xx10^(21)`
C. `1.00xx10^(22)`
D. `6.02xx10^(24)`

Answer

To find out how many unit cells are there in a 1.00 g cube-shaped ideal crystal of AB with a molecular weight of 60 g/mol and a NaCl-type lattice, we can follow these steps:

### Step 1: Determine the number of moles in 1.00 g of the crystal.
The molecular weight (M_W) of the compound AB is given as 60 g/mol. We can calculate the number of moles (n) in 1.00 g using the formula:

\[ n = \frac{\text{mass}}{\text{molecular weight}} = \frac{1.00 \text{ g}}{60 \text{ g/mol}} = \frac{1}{60} \text{ mol} \]

### Step 2: Calculate the number of formula units in 1.00 g.
Using Avogadro's number (N_A = \(6.022 \times 10^{23}\) units/mol), we can find the number of formula units (N) in the calculated moles:

\[ N = n \times N_A = \left(\frac{1}{60} \text{ mol}\right) \times (6.022 \times 10^{23} \text{ units/mol}) = \frac{6.022 \times 10^{23}}{60} \approx 1.0037 \times 10^{22} \text{ units} \]

### Step 3: Determine the number of unit cells.
In a NaCl-type lattice, the effective number of formula units (Z) per unit cell is 4. Therefore, to find the number of unit cells (N_cells), we can use the formula:

\[ N_{\text{cells}} = \frac{N}{Z} = \frac{1.0037 \times 10^{22} \text{ units}}{4} \approx 2.50925 \times 10^{21} \text{ unit cells} \]

### Final Answer:
Thus, the number of unit cells in a 1.00 g cube-shaped ideal crystal of AB is approximately \(2.51 \times 10^{21}\) unit cells.

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