A solid having density of `9 xx 10^(3) kg m^(-3)` forms face centred cubic crystals of edge length `200 sqrt2` pm. What is the molar mass of the solid ? [Avogadro constant `~= 6 xx 10^(23) mol^(-1), pi ~= 3`]

Explanation

To find the molar mass of the solid, we can use the formula for density in relation to the face-centered cubic (FCC) structure. The formula for density (D) is given by:

\[ D = \frac{Z \cdot M}{A^3 \cdot N_A} \]

Where:
- \(D\) = density of the solid (in kg/m³)
- \(Z\) = number of atoms per unit cell (for FCC, \(Z = 4\))
- \(M\) = molar mass of the solid (in kg/mol)
- \(A\) = edge length of the unit cell (in meters)
- \(N_A\) = Avogadro's number (in mol⁻¹)

### Step 1: Convert edge length from picometers to meters
The edge length \(A\) is given as \(200 \sqrt{2}\) pm. We need to convert this to meters.

\[ A = 200 \sqrt{2} \text{ pm} = 200 \sqrt{2} \times 10^{-12} \text{ m} \]

### Step 2: Substitute the values into the density formula
Given:
- Density \(D = 9 \times 10^3 \text{ kg/m}^3\)
- \(Z = 4\) (for FCC)
- \(N_A \approx 6 \times 10^{23} \text{ mol}^{-1}\)

Now substituting these values into the density formula:

\[ 9 \times 10^3 = \frac{4 \cdot M}{(200 \sqrt{2} \times 10^{-12})^3 \cdot (6 \times 10^{23})} \]

### Step 3: Calculate \(A^3\)
First, we need to calculate \(A^3\):

\[ A^3 = (200 \sqrt{2} \times 10^{-12})^3 = 200^3 \cdot (2^{3/2}) \cdot (10^{-12})^3 \]

Calculating \(200^3\):

\[ 200^3 = 8 \times 10^6 \]

Calculating \(2^{3/2}\):

\[ 2^{3/2} = 2 \sqrt{2} \approx 2.828 \]

Now combining these:

\[ A^3 \approx 8 \times 10^6 \cdot 2.828 \cdot 10^{-36} \approx 22.624 \times 10^{-30} \text{ m}^3 \]

### Step 4: Substitute \(A^3\) back into the density equation
Now substituting \(A^3\) back into the density equation:

\[ 9 \times 10^3 = \frac{4 \cdot M}{22.624 \times 10^{-30} \cdot (6 \times 10^{23})} \]

### Step 5: Solve for \(M\)
Rearranging the equation to solve for \(M\):

\[ M = \frac{9 \times 10^3 \cdot 22.624 \times 10^{-30} \cdot (6 \times 10^{23})}{4} \]

Calculating the right-hand side:

\[ M \approx \frac{9 \times 10^3 \cdot 22.624 \cdot 6}{4} \times 10^{-7} \approx \frac{1200.384}{4} \times 10^{-7} \approx 300.096 \times 10^{-7} \text{ kg/mol} \]

### Step 6: Convert \(M\) to grams per mole
Convert \(M\) from kg/mol to g/mol:

\[ M \approx 0.0300 \text{ kg/mol} = 30.0 \text{ g/mol} \]

### Final Answer
The molar mass of the solid is approximately \(30.0 \text{ g/mol}\).

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