A substance has a face centred cubic crystal with a density `1.984` g `cm^(-3)` and edge length 630 pm. Calculate the molar mass of the substance

Options

A. `98.63 g " mol"^(-1)`
B. `85.50 g " mol"^(-1)`
C. `74.70 g " mol"^(-1)`
D. `63.45 g " mol"^(-1)`

Explanation

To solve the problem, we need to calculate the molar mass of a substance that has a face-centered cubic (FCC) crystal structure, given its density and edge length. Here’s a step-by-step solution:

### Step 1: Understand the given data
- Density (ρ) = 1.984 g/cm³
- Edge length (a) = 630 pm = 630 x 10⁻¹² m = 630 x 10⁻⁸ cm = 6.30 x 10⁻⁸ cm

### Step 2: Determine the number of atoms per unit cell (Z) for FCC
For a face-centered cubic (FCC) structure, the number of atoms per unit cell (Z) is 4.

### Step 3: Use the formula for density
The formula for density in terms of molar mass (M), number of atoms per unit cell (Z), edge length (a), and Avogadro's number (Nₐ) is:
\[ \rho = \frac{Z \cdot M}{a^3 \cdot N_a} \]
Where:
- ρ = density
- Z = number of atoms per unit cell
- M = molar mass
- a = edge length
- Nₐ = Avogadro's number (6.022 x 10²³ mol⁻¹)

### Step 4: Rearrange the formula to solve for molar mass (M)
Rearranging the formula gives:
\[ M = \frac{\rho \cdot a^3 \cdot N_a}{Z} \]

### Step 5: Calculate a³ (the volume of the unit cell)
\[ a^3 = (6.30 \times 10^{-8} \text{ cm})^3 = 2.50 \times 10^{-23} \text{ cm}^3 \]

### Step 6: Substitute the values into the molar mass formula
Now substituting the known values into the rearranged formula:
\[ M = \frac{1.984 \, \text{g/cm}^3 \cdot 2.50 \times 10^{-23} \, \text{cm}^3 \cdot 6.022 \times 10^{23} \, \text{mol}^{-1}}{4} \]

### Step 7: Calculate M
Calculating the numerator:
\[ 1.984 \times 2.50 \times 6.022 = 29.73 \, \text{g/mol} \]
Now divide by 4:
\[ M = \frac{29.73}{4} = 7.43 \, \text{g/mol} \]

### Step 8: Final calculation
Now, we can calculate the final molar mass:
\[ M = 74.68 \, \text{g/mol} \]

### Conclusion
Thus, the molar mass of the substance is approximately **74.70 g/mol**.

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