Strontium oxide has a face centered cubic unit cell of oxide ions with the strontium ions in octahedral holes.If the radius of Sr²⁺ is 127 pm and the radius of O^2- is 135 pm,what is the density of strontium ions oxide? (100 cm = 1 × 10¹² pm) A) 0.209 g/cm³ B) 1.20 g/cm³ C) 2.39 g/cm³ D) 4 g/cm³ E) 4.78 g/cm³

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Quizplus · Accepted Answer

To calculate the density of strontium oxide (SrO), we first need to determine the formula unit mass and the volume of the unit cell. The face-centered cubic (fcc) unit cell of oxide ions means that each unit cell contains 4 oxide ions. Since strontium ions occupy the octahedral holes, there are also 4 strontium ions per unit cell, making the formula SrO.1. Calculate the formula unit mass of SrO:- Sr = 87.62 g/mol- O = 16.00 g/mol- SrO = 103.62 g/mol2. Determine the edge length (a) of the unit cell:Since the oxide ions form an fcc lattice and the radius of an O^2- ion is 135 pm, the edge length $a = 2\sqrt{2}r_{O^{2-}} = 2\sqrt{2} \times 135 \, \text{pm} = 382.68 \, \text{pm}$.3. Calculate the volume of the unit cell:$V = a^3 = (382.68 \, \text{pm})^3 = 5.59 \times 10^{7} \, \text{pm}^3$.Convert pm^3 to cm^3: $5.59 \times 10^{7} \, \text{pm}^3 = 5.59 \times 10^{-23} \, \text{cm}^3$.4. Calculate the number of formula units per unit cell:For SrO, there are 4 formula units per unit cell.5. Calculate the mass per unit cell:Mass = (number of formula units) $\times$ (molar mass) / (Avogadro's number)= $4 \times 103.62 \, \text{g/mol} / 6.022 \times 10^{23} \, \text{mol}^{-1}$= $6.89 \times 10^{-22} \, \text{g}$.6. Calculate the density:Density = mass / volume = $(6.89 \times 10^{-22} \, \text{g}) / (5.59 \times 10^{-23} \, \text{cm}^3)$ = 4.78 g/cm^3.Therefore, the correct answer is E) 4.78 g/cm^3.

Strontium Oxide Has a Face Centered Cubic Unit Cell of Oxide