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Chemistry for Engineers
Homework 3
Solutions and Solids
Answer Key
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Q1: 400g of the solvent acetone [(CH3)2CO] dissolved a 23 g ice [H2O] cube. Being
careful with units and significant figures…..
a) Calculate the solute molality and
b) Calculate the mole fraction of the solute
c) Calculate the volume of the solution (mL), given that it its molarity was
measured as 3.733 molL-1.
A: first work out number of moles of each:
molar mass acetone = 58.08 g/mol
moles acetone = 400/58.08 = 6.887 mol
molar mass water (H2O) = 18.016 g/mol
moles water (solute) = 23/18.016 = 1.2766 mol
a) Molality =
no. of moles H2O
mass acetone
= 1.2766/0.4
= 3.19 m or mol/kg
(3 + 3 + 3 marks)
1
Q1: 400g of the solvent acetone [(CH3)2CO] dissolved a 23 g ice [H2O] cube. Being
careful with units and significant figures…..
a) Calculate the solute molality and
b) Calculate the mole fraction of the solute
c) Calculate the volume of the solution (mL), given that it its molarity was
measured as 3.733 molL-1.
A:
b) Mole fraction =
no. of moles H2O
no. of moles acetone + no. moles H2O
= 1.2766/(6.887 + 1.2766) = 0.156 (no units)
just over 15% of molecules in the mixture are water molecules
c) volume (solution) =
no. of moles H2O
molarity H2O
= 1.2766/3.733
= 0.342 L or 342 mL
(3 + 3 + 3 marks)
Q2: By considering the colligative properties and possible ion pairing,
decide which of these solutions...
0.1 M Na2CO3
0.35 M HI
0.75 M glucose
0.15 M Al2(SO4)3
0.3 M urea
a) has the highest freezing point?
0.1 M Na2CO3
(particle concentration < 0.3 M
(ion pairing),
⇒ smallest depression of m.p.)
b) has the highest boiling point?
0.75 M glucose
(particle concentration = 0.75
M (no ion pairing)
⇒ greatest raising of b.p.)
c) exerts the lowest osmotic pressure? 0.1 M Na2CO3
(particle concentration < 0.3 M
(ion pairing),
⇒ small osmotic pressure)
(2 + 2 + 2 marks)
2
Q3: 13 g of FeBr2 was dissolved in 136 g water.
vapour pressure of water at 25 ºC is 23.76 mmHg
molal boiling point elevation constant for water Kb = 0.51 Kkgmol-1
a) What was the boiling point of the solution, assuming no ion pairing
occurs?
∆Tb = imsoluteKb
i = 3 if no ion pairs
molar mass (FeBr2) = 215.65 g/mol
msolute = moles(FeBr2)/mass water
moles(FeBr2) = mass FeBr2 /molar mass FeBr2 = 0.060283 mol
msolute = 0.060283 / 0.136
= 0.4433 molkg-1
∆Tb = imsoluteKb = 3 x 0.4433 x 0.51 = 0.678
b.p is 100.678 ºC
(4 + 5 + 5 marks)
Q3: 13 g of FeBr2 was dissolved in 136 g water.
vapour pressure of water at 25 ºC is 23.76 mmHg
molal boiling point elevation constant for water Kb = 0.51 Kkgmol-1
b) The vapour pressure of the solution was found to be 23.22 Torr.
Calculate the actual van’t Hoff factor, i
(5 marks)
Use Raoult’s Law for non-volatile solutes: Psoln = χsolv.P0solv
Psoln = 23.22 mmHg P0solv = 23.76 mmHg
χsolv = Psoln/P0solv = 23.22/23.76
= 0.97727
χparticles = 1 - 0.97727
= 0.022727
(4 + 5 + 5 marks)
3
Q3: 13 g of FeBr2 was dissolved in 136 g water.
vapour pressure of water at 25 ºC is 23.76 mmHg
molal boiling point elevation constant for water Kb = 0.51 Kkgmol -1
b) The vapour pressure of the solution was found to be 23.22 Torr.
Calculate the actual van’t Hoff factor, i
moles(H2O) = 136/18.016 = 7.54885 mol
moles(particles) = ( χparticles / χsolv ) x moles(H2O)
= (0.022727/0.97727) x 7.54885 = 0.17555
moles(FeBr2) x i = 0.17555
substitute moles (FeBr2) = 0.060283 mol (from part a))
i = 0.17555 / 0.060283 = 2.912
(4 + 5 + 5 marks)
Q3: c) The above FeBr2 solution (total volume 145 cm3) was placed
against a semi-permeable membrane. On the other side of the membrane
was placed a solution (270 cm3) containing a nitrate salt (13.63 g) for
which i = 1.92. Given that no net osmotic pressure resulted (Π = 0),
determine the molecular mass (and hence formula) of the nitrate
salt.
Π = iMRT
Π(FeBr2) = Π(nitrate)
iFeBr2MFeBr2 = initrateMnitrate
Mnitrate = iFeBr2MFeBr2 / initrate
MFeBr2 = moles/volume (L)
= 0.060277/0.145
= 0.4157 M
Mnitrate = 2.912 x 0.4157 / 1.92 = 0.63051 M
[nitrate] = mass/volume = 13.63/0.27 = 50.4815 gL-1
molar mass nitrate = 50.4815 / 0.63051 = 80.06 g/mol
atomic weight (NO3- ion) = 62.01
atomic weight (cation) = 80.06 - 62.01 = 18.05 (ammonium, NH4+)
(4 + 5 + 5 marks)
-unknown nitrate is NH4NO3
4
Q4: The ccp (face-centered cubic) unit cell of KBr has exactly the
same mass as the body-centered cubic unit cell of metal M. What is
metal M?
(6 marks)
A: First determine number of KBr formula units per
unit cell (cations are pink, anions are green)
KBr
KBr – formula units per unit cell
= [1 + (12 x ) + (8 x ) + (6 x )] = 4
cations
anions
mass = 4 x (39.10 + 79.90) = 476.00 Daltons/unit cell
for metal M: (bcc) count atoms per unit cell:
= 1 + (8 x ) = 2
mass of M atom = 476.0 Da / 2 = 238.00 Da
M is uranium (U)
Metal M
Q5: Unknown metal X crystallises in the non-close packed body-centred
cubic (bcc) structure. Its atomic radius is 219.5 pm and its density 3500
kgm-3
a) How many atoms are present per unit cell? 2
b) Given that the corner-to-corner distance (c) spans four atomic radii
(4r) use pythagoras to calculate the length (l) of each side of the cell.
from Pythagoras theorum:
l2 + l2 + l2 = c2
c = 4r (atoms touch each other)
3l2 = (4r)2 = 16r2
l = (16r2/3)½ sub r = 219.5 x 10-12 m
l = 506.9 pm
bcc
c) What is the volume of the unit cell?
v = (506.9 x 10-12)3
v = 1.30257 x 10-28 m3
(1 + 3 + 2 + 2 + 2 marks)
5
Q5: Unknown metal X crystallises in the non-close packed body-centred
cubic (bcc) structure. Its atomic radius is 219.5 pm and its density 3500
kgm-3
d) What is the mass of the unit cell?
mass unit cell = density x volume = 3500 x 1.30257 x 10-28 =
mass unit cell = 4.559 x 10-25 kg
e) Identify the metal.
For bcc, 2 atoms per unit cell, so:
bcc
10-25
mass atom = 2.2795 x
1 Da = 1.6605 x 10-27 kg
kg
mass atom = 137.28 Da
⇒ metal X is Ba
(1 + 3 + 2 + 2 + 2 marks)
6