LiBr have the density of
3
.
4
6
4
g
/
cm
3
,
rock salt structure, shown in figure
1
.
The atomic weight of Li
=
6
.
9
4
1
and Br
=
7
9
.
9
0
4
.
Caculate
(
a
)
the molar volumn of LiBr
(
b
)
the unit cell volumn
(
cm
3
/
unit
)

Question

LiBr have the density of
3
.
4
6
4
g
/
cm
3
,
 ﻿rock salt structure, shown in figure 
1
.
The atomic weight of Li
=
6
.
9
4
1
 ﻿and Br
=
7
9
.
9
0
4
.
Caculate 
(
a
)
the molar volumn of LiBr
(
b
)
the unit cell volumn
(
cm
3
/
unit
)

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Calculate the molar mass of LiBr**: To find the molar volume, we first need to calculate the molar mass of LiBr using the atomic weights of lithium (Li) and bromine (Br). ‖ ‖ M_{ ext{LiBr}} = M_{ ext{Li}} + M_{ ext{Br}} = 6.941 ext{ g/mol} + 79.904 ext{ g/mol} = 86.845 ext{ g/mol} ‖ ‖ 2 ⋮ **Calculate the molar volume of LiBr**: The molar volume is the volume occupied by one mole of a substance. We can calculate it using the density and the molar mass. ‖ ‖ V_{ ext{molar}} = \frac{M_{ ext{LiBr}}}{ ext{density}} = \frac{86.845 ext{ g/mol}}{3.464 ext{ g/cm}^3} = 25.07 ext{ cm}^3/ ext{mol} ‖ ‖ 3 ⋮ **Calculate the unit cell volume of LiBr**: The rock salt structure has a face-centered cubic (FCC) unit cell with 4 formula units per unit cell. The molar volume divided by Avogadro's number gives the volume of one formula unit, which we multiply by 4 to find the unit cell volume. ‖ ‖ V_{ ext{unit cell}} = \frac{V_{ ext{molar}}}{N_A} imes 4 = \frac{25.07 ext{ cm}^3/ ext{mol}}{6.022 imes 10^{23} ext{ formula units/mol}} imes 4 = 1.66 imes 10^{-22} ext{ cm}^3/ ext{unit cell} ‖ ∻1 Answer∻ ⚹ The molar volume of LiBr is 25.07 cm³/mol ⚹ ∻2 Answer∻ ⚹ The unit cell volume of LiBr is 1.66 × 10⁻²² cm³/unit cell ⚹ ∻Key Concept∻ ⚹ Molar volume is calculated using the molar mass and density of a substance. The unit cell volume in a crystal structure can be found by dividing the molar volume by Avogadro's number and multiplying by the number of formula units per unit cell. ⚹ ∻Explanation∻ ⚹ The molar volume is derived from the density and molar mass, providing the volume occupied by one mole of the substance. For crystalline substances with known structures, the unit cell volume can be calculated by relating the molar volume to the number of entities in a unit cell, using Avogadro's number to convert moles to individual formula units. ⚹◊What is the formula for calculating the molar volume of LiBr in a rock salt structure?⍭ Generate me a similar question◊

Sia