1. What is the volume of a face centered cubic unit cell made up of atoms having
ID: 999303 • Letter: 1
Question
1. What is the volume of a face centered cubic unit cell made up of atoms having a radius of 1.00 x 10-8 cm?
2. How many joules are required to raise the temperature of 7.5 g of water from 67.2 to 71.2 0C . Write your answer without units to three significant figures. Do not use scientific notation.
3. If you mx 200 mL of 0.050 M calcium hydroxide with 200 mL of 0.30 M HCl, what will be the concentration of calcium ion in the resulting solution?
4. A compound analyzes, by weight: P, 26.72%; N, 1209%; Cl, 61.17%. In a given experiment, 1.009 g of this compound was dissolved in 11.38 mL of benzene (d = 0.879 g/mL); the solution was observed to freeze at 437 0C. Previous measurements showed a freezing point of 5480C for the solvent used (C6H6; Kf = 5.120C/molal). What is the molecular formula of the compound in the benzene solvent?
5. Gold crystallizes in a b]fcc lattice with a unit-cell edge length of 4.08 angstrom. What is the radius of the Au based on this data to three significant figures in pm but do not include units with your answer. Just the number value. Do not use scientific notation.
Explanation / Answer
1) For FCC,
Radius = a / 2(2)1/2
a= edge length
a = Radius (2(2)1/2)
a = 10^-8 X 2 X 1.414
a = 2.828 X 10^-8 cm
Volume =a^3 = 22.617 X 10^-24 cm^3
2) The specific heat of water = 4.184 J / g 0C
Heat absorbed = Mass of water X specific heat X change in temperature
Heat = 7.5 X 4.184 X (71.2-67.2) = 125.52 Joules
3) Moles of Calcium ion added = Moles of Ca(OH)2 = Molarity X volume = 0.05 X 0.2 = 0.01
Total volume = 200mL + 200mL = 400mL
New concentration = Moles / Volume in litres = 0.01 / 0.4 = 0.025 M
4) The molecular weight of compound can be determined from depression in freezing point
Depression in freezing point = Kf X molality of the compound
(5.48 - 4.37 ) = 5.12 X molality
Molality = 0.217
Molality = Mass of compound / molecular weight x mass of benzene in Kg
Mass of benzene = Density x volume = 0.879 X 11.38 = 10 grams = 0.01 Kg
0.217 = 1.009 / Mol wt X 0.01
Mol wt = 464.97 grams / mole
% P 26.72 so mass of Phosphorous in compound = 0.2672 X 464.97 = 124.24 grams
Moles = Mass / At wt = 124.24 / 31 = 4
% nitrogen = 12.09 so mass of nitrogen = 0.1209 X 464.97 = 56.21 grams
Moles = Mass / at wt = 56.21 / 14 = 4
% of Cl = 61.17
Mass = 284.42
Moles = 284.42 / 35.5 = 8
Possible molecular formula = P4N4Cl8
5) Edge length = 4.08 angstrom
a = 4R/31/2
Radius = a X 31/2 / 4
Radius = 4.08 X 1.732 / 4 = 1.7667 angstrom = 176.67 pm
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.