The density of CsCl should be calculated. Concept introduction: As the caesium has a body centred cubic (bcc) crystal structure, nine atoms are associated with a bcc unit cell. One atom is located at each of the eight corners of the cube and one at the centre. The three atoms along a diagonal through the cube are in contact. The length of the cube diagonal (the distance from the farthest upper-right corner to the nearest lower left corner) is four times the atomic radius. Also, shown below is the fact the diagonal of a cube is equal to 3 × l . The length of an edge, l, is what is given. The right triangle must conform to the Pythagorean formula a 2 = b 2 + c 2 . From the Pythagorean formula, following the relationship between r and l can obtain. l 3 = 2 ( r Cs + + r Cl - ) From the above equation, volume ( v ) of the unit cell can be calculated as by the following equation as this is a cube. v = l 3 v = ( 2 ( r C s + + r C l − ) 3 ) 3 Density ( d ) defines as the mass per unit volume. d = m v The following relationship is used to find the mass of a unit cell. m = n M CsCl N A Here, n is molecules per unit cell . N A is avagadro constant M CsCl is relative molar mass of CsCl
The density of CsCl should be calculated. Concept introduction: As the caesium has a body centred cubic (bcc) crystal structure, nine atoms are associated with a bcc unit cell. One atom is located at each of the eight corners of the cube and one at the centre. The three atoms along a diagonal through the cube are in contact. The length of the cube diagonal (the distance from the farthest upper-right corner to the nearest lower left corner) is four times the atomic radius. Also, shown below is the fact the diagonal of a cube is equal to 3 × l . The length of an edge, l, is what is given. The right triangle must conform to the Pythagorean formula a 2 = b 2 + c 2 . From the Pythagorean formula, following the relationship between r and l can obtain. l 3 = 2 ( r Cs + + r Cl - ) From the above equation, volume ( v ) of the unit cell can be calculated as by the following equation as this is a cube. v = l 3 v = ( 2 ( r C s + + r C l − ) 3 ) 3 Density ( d ) defines as the mass per unit volume. d = m v The following relationship is used to find the mass of a unit cell. m = n M CsCl N A Here, n is molecules per unit cell . N A is avagadro constant M CsCl is relative molar mass of CsCl
Solution Summary: The author explains how the density of CsCl should be calculated.
As the caesium has a body centred cubic (bcc) crystal structure, nine atoms are associated with a bcc unit cell. One atom is located at each of the eight corners of the cube and one at the centre. The three atoms along a diagonal through the cube are in contact. The length of the cube diagonal (the distance from the farthest upper-right corner to the nearest lower left corner) is four times the atomic radius. Also, shown below is the fact the diagonal of a cube is equal to 3×l. The length of an edge, l, is what is given.
The right triangle must conform to the Pythagorean formula a2=b2+c2.
From the Pythagorean formula, following the relationship between r and l can obtain. l3=2(rCs++rCl-)
From the above equation, volume ( v ) of the unit cell can be calculated as by the following equation as this is a cube.
v=l3
v=(2(rCs++rCl−)3)3
Density ( d ) defines as the mass per unit volume.
d=mv
The following relationship is used to find the mass of a unit cell.
m=nMCsClNA
Here,
nismoleculesperunitcell.NAis avagadroconstantMCsClis relative molar mass of CsCl
Chemistry
4(a)For each of the following substances describe the importance of dipole-dipole interactions and hydrogen bonding for their dissolution in H2O: (a) Br2, (b) CH4.(b)Sketch an enthalpy diagram for solution formation of a solid in liquid and identify the changes in the mixing steps.
3. What is the normal melting point of the substance?4. What phase(2) will exist at 1 atm and 70 degree Celcius?
Use these data to draw a qualitative phase diagram for eth-ylene (C₂H₄ ). Is C₂H₄ (s) more or less dense than C₂H₄ (l)?bp at 1 atm: -103.7°C ; mp at 1 atm: -169.16°C ;Critical point:9.9°C and 50.5 atm ;Triple point: -169.17°C and 1.20X10^-3atm
Chapter 12 Solutions
General Chemistry: Principles and Modern Applications (11th Edition)