$\mathrm{LiBr}$ have the density of $3.464 \mathrm{~g} / \mathrm{cm} 3$, rock salt structure, shown in figure 1 . The atomic weight of $\mathrm{Li}$ $=6.941$ and $\mathrm{Br}=79.904$. Caculate (a)the molar volumn of $\mathrm{LiBr}(\mathrm{b})$ the unit cell volumn(cm3/unit)

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∻Solution∻ ‖ 1 ⋮ **Calculate the molar mass of LiBr**: The molar mass of $LiBr$ is the sum of the atomic weights of $Li$ and $Br$. $$ \text{Molar mass of } LiBr = 6.941 + 79.904 = 86.845 \, \text{g/mol} $$ ‖ ‖ 2 ⋮ **Calculate the molar volume of LiBr**: The molar volume is given by the formula: $$ \text{Molar volume} = \frac{\text{Molar mass}}{\text{Density}} $$ $$ \text{Molar volume} = \frac{86.845 \, \text{g/mol}}{3.464 \, \text{g/cm}^3} = 25.07 \, \text{cm}^3/\text{mol} $$ ‖ ‖ 3 ⋮ **Determine the unit cell volume**: LiBr has a rock salt structure, which is a face-centered cubic (FCC) structure. In an FCC structure, there are 4 formula units per unit cell. The unit cell volume can be calculated as: $$ \text{Unit cell volume} = \frac{\text{Molar volume}}{\text{Number of formula units per unit cell}} $$ $$ \text{Unit cell volume} = \frac{25.07 \, \text{cm}^3/\text{mol}}{4} = 6.2675 \, \text{cm}^3/\text{unit cell} $$ ‖ ∻Answer∻ ⚹ The molar volume of $LiBr$ is $25.07 \, \text{cm}^3/\text{mol}$ and the unit cell volume is $6.2675 \, \text{cm}^3/\text{unit cell}$. ⚹ ∻Key Concept∻ ⚹ Molar volume and unit cell volume calculations ⚹ ∻Explanation∻ ⚹ The molar volume is calculated by dividing the molar mass by the density. The unit cell volume is then determined by dividing the molar volume by the number of formula units in the unit cell for the given crystal structure. ⚹◊What is the formula to calculate the molar volume of a compound in a rock salt structure?⍭ Generate me a similar question◊

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