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Chemistry
Posted about 1 month ago

LiBr\mathrm{LiBr} have the density of 3.464 g/cm33.464 \mathrm{~g} / \mathrm{cm} 3, rock salt structure, shown in figure 1 . The atomic weight of Li\mathrm{Li} =6.941=6.941 and Br=79.904\mathrm{Br}=79.904. Caculate (a)the molar volumn of LiBr(b)\mathrm{LiBr}(\mathrm{b}) the unit cell volumn(cm3/unit)
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Answer from Sia
Posted about 1 month ago
Solution
1
Calculate the molar mass of LiBr: The molar mass of LiBrLiBr is the sum of the atomic weights of LiLi and BrBr. Molar mass of LiBr=6.941+79.904=86.845g/mol \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: Molar volume=Molar massDensity \text{Molar volume} = \frac{\text{Molar mass}}{\text{Density}} Molar volume=86.845g/mol3.464g/cm3=25.07cm3/mol \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: Unit cell volume=Molar volumeNumber of formula units per unit cell \text{Unit cell volume} = \frac{\text{Molar volume}}{\text{Number of formula units per unit cell}} Unit cell volume=25.07cm3/mol4=6.2675cm3/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 LiBrLiBr is 25.07cm3/mol25.07 \, \text{cm}^3/\text{mol} and the unit cell volume is 6.2675cm3/unit cell6.2675 \, \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.

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