DUE Jan 31, 2001 Computer homework: Solutions 1-2

1. Chromium crystallizes in a body centered cubic lattice. The planes give the strongest reflections for a spacing d = 2.492 A. What is the angle for the first order diffraction of X-rays from a molybdenum X-ray source l = 0.712 A?

2. The unit cell for a pure xenon fluoride is shown to the right. What is the formula for the compound?

Xe = 1/8 x 8 = 1

1 x 1 = 1 à 2 Xe

F = 1/4 x 8 = 2

1 x 2 = 2 à 4 F

Formula = |__XeF₂________
 
 

3. Europium crystallizes in a body-centered cubic lattice. The density of Eu is 5.26 g/cm3.

1/8 x 8 + 1 x 1 = 2 atoms/unit cell

a. What is the length of the edge of the unit cell?

D = M of unit cell / volume of unit cell = 151.97 g/mol (1 mol/6.022 x 10²³ atoms) x 2 atoms/unit cell / e³ = 5.26 g/cm³

e³ = (151.97 x 2)/ (6.022 z 10²³ x 5.26) cm³ = 9.595 x 10^-23; e = 4.58 x 10^-8 cm

Edge = |_4.58 x 10^-8 cm = 4.58 A = 0.458 nm_

b. What is the radius of the europium atom?

For bcc, body diagonal D = (3)^0.5 e = 4 R

R = (3)^0.5 (0.458 nm) / 4 = 0.198 nm = 1.98 A = 1.98 x 10^-8 cm

Radius = | 0.198 nm = 1.98 A = 1.98 x 10^-8 cm

4. KCl has the same structure as NaCl. The length of the unit cell is 628 pm. Calculate the density (g/cm3) of the KCl.

K = 1/4 x 12 + 1 x 1 = 4 K⁺

Cl = 1/8 x 8 + 1/2 x 6 = 4 Cl^- per unit cell

D = M/V = [(39.10 + 35.45) g/mol x (1 mol / 6.022 x 10²³ ions x 4 ion units / unit cell] / (6.28 x 10^-8 cm)³

= 4.952 x 10^-22 / 2.477 g/cm³ = 1.999 = 2.00 g/cm³

Density = |_2.00 g cm³ ___

5. Use the following properties to decide which type of solid corresponds to each of the following chemical species.

a. Melting point 782oC; hard, brittle solid covalent network ionic metallic molecular

Conductor only when melted

b. Melting point -140oC; soft solid covalent network ionic metallic molecular

Non-conductor when melted
 
 

6. Circle the chemical species in each of the following pairs which would be expected to be more soluble in water. Write a word or two below each pair to explain your choice. (dipole, polarizability, etc.)

7. The solubility of carbon dioxide in water at 20oC and 1.00 atm is 0.161g in 100. ml. A soft drink is carbonated with carbon dioxide at 5.50 atm . What is the solubility of of carbon dioxide at this temperature

C₁/P₁ = C₂/P₂ = 0.161 g/100 mL / 1.00 atm = C₂ / 5.50 atm.

C₂ = 0.161 g/100 mL X 5.50 atm/1.0 atm = .886 g/100 mL

Solubility = _.886 g / 100 mL_
 
 

8. The heat of solution of NaCl at 298 K for the preparation of very dilute aqueous solution is 3.9 kJ/mol. The lattice energy of NaCl is 789 kJ/mol. What is the heat of hydration of the ions of this compound?

The heat of solution is positive which means that more energy was put into breaking the lattice into ions than was regained by hydrating the ions. The Heat of hydration is 3.9 - 789 = -785 kJ/mol

/\Hhydration = |_-785 kJ/mol_____
 
 

9. A solution of NaCl in water has a concentration of 19.5 % by Wt. Calculate the molality of the solution.

That means that if a sample of 100 g is assumed there would be 19.5 g of NaCl and 81.5 g of water.

Molality = moles of solute/kg of solvent = (19.5 g/ 58.44 g/mol) / (80.5 g/1000 g/kg) = 0.3337mol/0.0805 kg =4.145 m

m= _4.145 m _______

10. The violet solution containing 28.00% by mass Cr₂(SO₄)₃ (MM = 392.2) in water (MM=18.02) has a density of 1.3325g/ml, calculate 1.3325 g/mL x 1000 mL/L = 1332.5 g/L solution; 1332.5 x 28.00/100 = 373.1 g salt; 1332.5 - 373.1 = 959.4 g H₂O

m = moles of solute/kg of solvent = (373.1 g / 392.2 g/mol) /(959.4 g / 1000 g/kg) = 0.9513 mol / 0.9594 kg = 0.9916 m

m = |_0.9916 m______

b. Molarity of Cr₂(SO₄)₃

M = moles of solute / liter of solution = 0.9513 mol / L (from above)

M = |_0.9513 M______

c. molfraction of Cr₂(SO₄)₃

C = mol of salt / total mol of soln = 0.9513 mol / (0.9513 + 959.4 g/18.02 g/mol) = 0.9513 / (0.9513 + 53.241) = 0.9513/54.192

= 0.01755

X = |_0.01755______