CH.11
Condensation occurs when?
When:
ΔH>0
ΔS>0
How many atoms are in a simple cubic unit cell?bbc
corners:8
8*1/8
Net:1
What is the formula for molarity?
Moles of solute/Liters of Solution
What are the 4 Colligative Properties?
–Vapor-pressure lowering
–Boiling-point elevation
–Freezing-point depression
–Osmotic pressure
If ΔH is negative is it exothermic or endothermic?
Exothermic
How many atoms are in a body-centered cubic packing cell?
corners:8 center:1
8*1/8+1
Net:2
What is the formula for molality?
Moles of solute/ Mass of solvent(kg)
The vapor pressure of an aqueous solution is found to be 24.90 mmHg at 25 °C. What is the mole fraction of solute in this solution? The vapor pressure of water is 25.756 mm Hg at 25 °C.
Use Raoult's Law:
Psolution = (χsolv) (Psolvo)
24.90 = (x) (25.756)
x = 0.966765 (this is the solvent mole fraction)
χsolute = 1 − 0.966765 = 0.033235
What is the triple point?
A point at which all three phases coexist in equilibrium.
How many atoms are in a face-centered cubic unit cell?
Net:4
one-eighth of an atom at each of the eight corners (8×18=1 atom from the corners) and one-half of an atom on each of the six faces (6×12=3 atoms from the faces)
A solution of H2SO4 with a molal concentration of 8.010 m has a density of 1.354 g/mL. What is the molar concentration of this solution?
8.010 m means 8.010 mol / 1 kg of solvent
(8.010 mol) (98.0768 g/mol) = 785.6 g of solute
785.6 g + 1000 g = 1785.6 g total for solute and solvent in the 8.010 m solution.
1785.6 g / 1.354 g/mL = 1318.76 mL
8.01 moles / 1.31876 L = 6.0739 M
A solution is composed of 1.40 mol cyclohexane (Pcyo = 97.6 torr) and 2.50 mol acetone (Paco = 229.5 torr). What is Psolution, the total vapor pressure, above this solution?
1) Calculate the mole fraction of each substance:
cyclohexane ⇒ 1.40 mol / (1.40 mol + 2.50 mol) = 0.358974359
acetone ⇒ 2.50 mol / (1.40 mol + 2.50 mol) = 0.641025641
2) Use Raoult's Law:
Psolution = (Pcyo) (χcy) + (Paco) (χac)
Psolution = (97.6 torr) (0.358974359) + (229.5 torr) (0.641025641)
Psolution = 35.03589744 + 147.1153846 = 182.15 torr
Determine ΔHvap for a compound that has a measured vapor pressure of 24.3 torr at 273 K and 135 torr at 325 K.
ln (135/24.3) = (x / 8.31447) (1/273 minus 1/325)
1.7148 = (x / 8.31447) (0.00058608)
1.7148 = 0.000070489x
x = 24327 J/mol = 24.3 kJ/mol
Polonium metal crystallizes in a simple cubic arrangement, with the edge of a unit cell having a length d = 334 pm. What is the radius in picometers of a polonium atom?
d = 2r. Solving for r, we get r = d / 2. Hence, the radius of a polonium atom is 334 pm / 2 =
167 pm.
What is the molarity of a solution that was prepared by dissolving 82.0 g of CaCl2 (molar mass = 111.1 g/mol) in enough water to make 812 mL of solution?
0.909 M
Camphor (C6H16O) melts at 179.8 °C, and it has a particularly large freezing point depression constant, Kf = 40.0 °C/m. When 0.186 g of an organic substance of unknown molar mass is dissolved in 22.01 g of liquid camphor, the freezing point of the mixture is found to be 176.7 °C. What is the molar mass of the solute?
179.8 − 176.7 = 3.1 °C
Δt = i Kf m
3.1 °C = (1) (40.0 °C kg mol¯1) (x / 0.02201 kg)
3.1 °C = (1) (1817.356 °C mol¯1) (x)
x = 0.001705775 mol
A certain liquid has a vapor pressure of 6.91 mmHg at 0 °C. If this liquid has a normal boiling point of 105 °C, what is the liquid's heat of vaporization in kJ/mol?
ln (6.91/ 760) = (x / 8.31447) (1/378.15 minus 1/273.15)
-4.70035 = (x / 8.31447) (-0.0010165)
4.70035 = 0.00012226x (notice I got rid of the negative signs)
x = 38445 J/mol = 38.4 kJ/mol
Nickel crystallizes in a face-centered cubic lattice. If the density of the metal is 8.908 g/cm3, what is the unit cell edge length in pm?
This problem is like the one above, it just stops short of determining the atomic radius.
1) Calculate the average mass of one atom of Ni:
58.6934 g mol¯1 / 6.022 x 1023 atoms mol¯1 = 9.746496 x 10¯23 g/atom
2) Calculate the mass of the 4 nickel atoms in the face-centered cubic unit cell:
(9.746496 x 10¯23 g/atom) (4 atoms/unit cell) = 3.898598 x 10¯22 g/unit cell
3) Use density to get the volume of the unit cell:
3.898598 x 10¯22 g / 8.908 g/cm3 = 4.376514 x 10¯23 cm3
4) Determine the edge length of the unit cell:
4.376514 x 10¯23cm33 = 3.524 x 10¯8 cm
5) Convert cm to pm:
cm = 10¯2 m; pm = 10¯12 m.
Consequently, there are 1010 pm/cm
(3.524 x 10¯8 cm) (1010 pm/cm) = 352.4 pm
Calculate the molality of a 3.5M KCL solution that has a density of 1.05g/ml?
.
4.4m
Calculate the osmotic pressure of a 0.25 M aqueous solution of sucrose, C12H22O11, at 37 C
= (0.25 mol L-1 ) × (0.08206 L atm K-1 mol-1 ) × ((37 + 273) K) = 6.4 atm