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Unit cell of the iron crystal has ed...

Unit cell of the iron crystal has edge length of 288 pm and density of 7.86 g `cm^(-3)`. Determinte the type of crystal lattice. Atomic mass of Fe= 56 g `mol^(-1)`

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To determine the type of crystal lattice for iron given the edge length and density, we can follow these steps: ### Step 1: Gather the Given Data - Edge length of the unit cell (a) = 288 pm = 288 × 10^(-12) m = 2.88 × 10^(-8) cm - Density (d) = 7.86 g/cm³ - Atomic mass of Fe (m) = 56 g/mol ### Step 2: Write the Formula for Density The formula for density (d) in terms of the number of atoms in the unit cell (z), molar mass (m), Avogadro's number (N_A), and volume (V) is given by: \[ d = \frac{z \cdot m}{N_A \cdot V} \] For a cubic unit cell, the volume (V) can be expressed as: \[ V = a^3 \] Thus, the density formula becomes: \[ d = \frac{z \cdot m}{N_A \cdot a^3} \] ### Step 3: Rearrange the Formula to Solve for z Rearranging the formula to solve for z gives us: \[ z = \frac{d \cdot N_A \cdot a^3}{m} \] ### Step 4: Substitute the Known Values Now, we can substitute the known values into the equation. We need to use Avogadro's number: \[ N_A = 6.022 \times 10^{23} \text{ mol}^{-1} \] Substituting the values: - d = 7.86 g/cm³ - N_A = 6.022 × 10^{23} mol⁻¹ - a = 2.88 × 10^(-8) cm - m = 56 g/mol Calculating \( a^3 \): \[ a^3 = (2.88 \times 10^{-8})^3 = 2.39 \times 10^{-23} \text{ cm}^3 \] Now substituting into the equation for z: \[ z = \frac{7.86 \cdot (6.022 \times 10^{23}) \cdot (2.39 \times 10^{-23})}{56} \] ### Step 5: Calculate z Calculating the numerator: \[ 7.86 \cdot 6.022 \times 10^{23} \cdot 2.39 \times 10^{-23} \approx 7.86 \cdot 6.022 \cdot 2.39 \approx 111.4 \] Now divide by 56: \[ z \approx \frac{111.4}{56} \approx 1.99 \] ### Step 6: Determine the Type of Crystal Lattice The value of z indicates the number of atoms per unit cell: - If z = 1, it is Simple Cubic (SC). - If z = 2, it is Body-Centered Cubic (BCC). - If z = 4, it is Face-Centered Cubic (FCC). Since z ≈ 2, we conclude that the crystal lattice of iron is **Body-Centered Cubic (BCC)**. ### Summary The type of crystal lattice for iron is BCC. ---

To determine the type of crystal lattice for iron given the edge length and density, we can follow these steps: ### Step 1: Gather the Given Data - Edge length of the unit cell (a) = 288 pm = 288 × 10^(-12) m = 2.88 × 10^(-8) cm - Density (d) = 7.86 g/cm³ - Atomic mass of Fe (m) = 56 g/mol ### Step 2: Write the Formula for Density ...
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