Version 001 – Calculating Concentrations WKST – vanden bout – (51165)
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before answering.
ChemPrin3e G 13
001 (part 1 of 2) 10.0 points
Determine the mass of anhydrous copper(II)
sulfate that must be used to prepare 250 mL
of 1.9 M CuSO4 (aq).
Correct answer: 75.8148 g.
Explanation:
V = 250 mL = 0.25 L
M = 1.9 M
FWCuSO4 = 63.546 g/mol
+ 32.066 g/mol
+ 4 (15.9994 g/mol)
= 159.61 g/mol
mCuSO4 = (1.9 M) (0.25 L) (159.61 g/mol)
= 75.8148 g
003 10.0 points
A chemist studying the properties of photographic emulsions needed to prepare 500 mL
of 0.178 M AgNO3 (aq). What mass of silver
nitrate must be placed into a 500 mL volumetric flask, dissolved, and diluted to the mark
with water?
Correct answer: 15.1187 g.
Explanation:
v = 500 mL = 0.5 L
mAgNO3 = (0.178 M) (0.5 L)
× (169.873 g/mol)
= 15.1187 g
Brodbelt 013 015
004 10.0 points
How much NaNO3 is needed to prepare 225
mL of a 1.55 M solution of NaNO3 ?
1. 4.10 g
3. 12.3 g
Correct answer: 118.603 g.
4. 0.132 g
FWCuSO4 ·5 H2 O = FWCuSO4 + 5 (FWH2O )
= 159.61 g/mol
+ 10 (1.0079 g/mol)
+ 5 (15.9994 g/mol)
= 249.69 g/mol
mCuSO4 ·5 H2 O = (1.9 M) (0.25 L) (249.69 g/mol)
= 118.603 g
ChemPrin3e G 05
M = 0.178 M
FWAgNO3 = 107.8682 g/mol + 14.0067 g/mol
+ 3 (15.9994 g/mol)
= 169.873 g/mol
002 (part 2 of 2) 10.0 points
Determine the mass of CuSO4 · 5 H2O that
must be used to prepare 250 mL of 1.9 M
CuSO4 (aq).
Explanation:
1
2. 29.6 g correct
5. 0.244 g
Explanation:
V = 225 mL
M = 1.55 M
1 L soln
1000 mL
1.55 mol NaNO3
×
1 L soln
85 g NaNO3
×
1 mol NaNO3
= 29.6 g NaNO3
? g NaNO3 = 225 mL ×
Brodbelt 03 13
Version 001 – Calculating Concentrations WKST – vanden bout – (51165)
005 10.0 points
How many moles of HCl are present in 40.0
mL of a 0.035 M solution?
1. 0.012 mol
2. 0.0060 mol
3. 0.25 mol
in the two individual solutions:
? mol Ca(OH)2 (soln 1)
= 0.255 L soln
0.250 mol Ca(OH)2
×
1 L soln
= 0.06375 mol Ca(OH)2
? mol Ca(OH)2 (soln 2)
= 0.055 L soln
0.65 mol Ca(OH)2
×
1 L soln
= 0.03575 mol Ca(OH)2
4. 0.0012 mol
5. 0.0014 mol correct
Explanation:
V = 40.0 mL
2
M = 0.035 M
? mol HCl = 40.0 mL soln
1L
0.035 mol HCl
×
×
1000 mL
1 L soln
= 0.0014 mol HCl
Total mol Ca(OH)2 = 0.06375 mol
+ 0.03575 mol
= 0.0995 mol Ca(OH)2
The total volume of the final solution is the
sum of the volumes of the individual solutions.
Total L = 0.255 L + 0.055 L = 0.31 L soln
Molarity is moles solute per L of solution.
0.0995 mol Ca(OH)2
0.31 L soln
= 0.321 M Ca(OH)2
? M Ca(OH)2 =
Brodbelt 013 515
006 10.0 points
What is the final concentration of Ca(OH)2
when 255 mL of 0.250 M Ca(OH)2 is mixed
with 55.0 mL of 0.65 M Ca(OH)2 ?
1. 0.390 M
Msci 14 0813
007 10.0 points
What is the effective molality of a solution
containing 12.0 g of KI in 550 g water? Assume 100 percent ionic dissociation.
2. 0.642 M
1. 0.072 molal
3. 0.780 M
2. 0.26 molal correct
4. 2.90 M
3. 0.42 molal
5. 0.900 M
4. 0.066 molal
6. 0.321 M correct
5. 0.13 molal
Explanation:
V1 Ca(OH)2 = 255 mL [Ca(OH)2 ]1 = 0.250 M
[Ca(OH)2]2 = 0.65 M
V2 Ca(OH)2 = 55 mL
The total moles of Ca(OH)2 in the final
solution will be the sum of the moles present
6. 0.59 molal
Explanation:
mKI = 12.0 g
mH2 O = 550 g
Because KI is a 1:1 salt, you get one cation
and one anion for every single formula unit
Version 001 – Calculating Concentrations WKST – vanden bout – (51165)
3
that dissolves. Therefore, you’ll get a DOUBLING of the stated molality for the effective
molality. The formula weight of KI is 166
g/mol, so the number of moles of KI is
1 mol KI 12.0 g KI
= 0.0723 mol KI ,
166 g KI
Holt da 13 1 sample 1
010 10.0 points
What is the molarity of a 3.047 L solution
that is made from 11.29 g of NaCl?
and the (stated) molality of the solution
would then be
0.0723 mol KI
= 0.131 m .
0.550 kg H2 O
Explanation:
Vsolution = 3.047 L
M =?
But recall that the effective molality would be
twice the stated molality here, so the effective
molality is 0.262 m.
Nlib 11 0020
008 10.0 points
If you mix 3 moles of ethylene glycol (antifreeze) in 4165 grams of water, what is the
molality of the solution?
Correct answer: 0.0634032 M.
mol solute
L soln
? mol NaCl = 11.29 g NaCl
1 mol NaCl
×
58.44 g NaCl
= 0.19319 mol NaCl
M=
0.19319 mol
mol NaCl
=
L soln
3.047 L
= 0.0634032 M
? M=
Correct answer: 0.720288 m.
Explanation:
nethylene glycol = 3 mol
mwater = 4165 g
Molality (m) is moles solute per kilogram
of solvent. The solute is ethylene glycol. The
solvent is water and 4165 g = 4.165 kg.
?m=
3 mol ethylene glycol
= 0.720288 m
4.165 kg H2 O
Nsci 14 0808
009 10.0 points
1.9 g of NaCl and 6.1 g of KBr were dissolved
in 48 g of water. What is the molality of NaCl
in the solution?
Molality 08 44a
011 10.0 points
Calculate the molality of sucrose in a solution
composed of 11.31 g of sucrose (C12 H22 O11 )
dissolved in 606 mL of water.
Correct answer: 0.054524 m.
Explanation:
mC12 H22 O11 = 11.31 g
VH2 O = 606 mL = 0.606 L
MWC12 H22 O11 = 342.296 g/mol
Correct answer: 0.6773 m.
Explanation:
mNaCl = 1.9 g
mwater = 48 g
mH2 O
mKBr = 6.1 g
mol NaCl
kg water
1.9 g
58.4 g · mol−1 NaCl
=
0.048 kg water
= 0.6773 m
mNaCl =
mNaCl = 11.29 g
1 kg
= (0.606 L)
1L
= 0.606 kg
Thus the molality is
moles solute
kg solvent
11.31 g sucrose
342.296 g/mol sucrose
=
0.606 kg H2 O
= 0.054524 m
mC12 H22 O11 =
Version 001 – Calculating Concentrations WKST – vanden bout – (51165)
Msci 14 0809
012 10.0 points
What additional information, if any, would
enable you to calculate the molality of a 7.35
molar solution of a nonelectrolyte solid dissolved in water?
1. Both the density of the solution and
the molecular weight of the solute would be
needed. correct
2. Only the density of the solution would be
needed.
3. Only the molecular weight of the solute
would be needed.
4. Only the density of water would be
needed.
5. None is needed.
Explanation:
mol solute
L solution
mol solute
molality =
kg solvent
molarity =
The density of the solution can be used to
convert volume (1 L) of solution into mass of
solution. Then the molecular weight of the
solute (given or calculated from the formula)
can be used to convert the number of moles
solute in 1 L solution into mass of solute
in grams. The mass of the solvent is the
difference between the mass of the solution
and the mass of the solute (both of which
have been calculated). Substitute the values
into the molality formula and calculate.
Molality 08 46c
013 10.0 points
Calculate the molality of 10.5 M NH3 (aq)
with a density of 0.9344 g/cm3 .
Correct answer: 13.8966 m.
Explanation:
MW = 17.0305 g/mol
M = 10.5 M
4
d = 0.9344 g/cm3
Assume 1 L of 10.5 M NH3 (aq); it will
contain 10.5 mol NH3 with a mass of
(10.5 mol)(17.0305 g/mol) = 178.82 g .
The density of the 1 L of solution is
0.9344 g/cm3 ·
1000 cm3
= 934.4 g/L ,
1L
so the total mass of the solution is 934.4 g,
which leaves 934.4 g − 178.82 g = 755.58 g of
water.
Therefore, the molality is
m=
10.5 mol NH3
= 13.8966 m .
0.75558 kg solvent
Nsci 14 0803
014 10.0 points
Formalin is a solution of 40.0% formaldehyde
(H2 CO), 10.0% methyl alcohol (CH3 OH), and
50.0% water by mass. Calculate the mole
fraction of methyl alcohol in formalin.
Correct answer: 0.0706436.
Explanation:
In a 100 g formalin, solution the masses are
formaldehyde 40.0 g, methyl alcohol 10.0 g,
and water 50.0 g.
ntotal = nCH3 OH + nH2 CO + nwater
10 g
40 g
=
+
32 g/mol 30 g/mol
50 g
+
18 g/mol
= 4.42361 mol
The mole fraction of methyl alcohol is
nCH3 OH
ntotal
10.0 g
32.0 g/mol
=
4.42361 mol
= 0.0706436
X=
Nsci 14 0801
015 10.0 points
Toluene (C6 H5 CH3 ) is a liquid compound
Version 001 – Calculating Concentrations WKST – vanden bout – (51165)
similar to benzene (C6 H6 ). Calculate the
mole fraction of toluene in the solution that
contains 112 g toluene and 72.0 g benzene.
Correct answer: 0.568.
Explanation:
mtoluene = 112 g
ntoulene = (112 g toluene)
mbenzene = 72.0 g
1 mol 92.14 g
= 1.21 mol
nbenzene = (72.0 g benzene)
1 mol 78.11 g
= 0.922 mol
The total number of moles of all species
present is
1.21 mol + 0.922 mol = 2.13 mol
The mole fraction of toluene is then
Xtoluene =
1.21 mol
ntoluene
= 0.568
=
ntotal
2.13 mol
Msci 14 0802
016 10.0 points
The mole fraction of a certain nonelectrolyte
compound in a solution containing only that
substance and water is 0.100. The molecular
weight of water is 18.0 g/mol. What additional information is needed to determine the
molality of the solution?
1. The density of the solute.
2. The density of the solution.
3. The molecular weight of the compound.
4. The mole fraction of water in the solution.
5. No additional information; the molality
can be calculated from the information given.
correct
Explanation:
Here we can assume that we have 1 mol
total. (In fact, we can choose any number of
moles, but the math is easier if you choose 1
mol.) If the mole fraction of the substance
5
is 0.100, you can then assume that you have
0.100 mol of the substance, and the remaining
0.900 mol is H2 O. The molality of a solution
is determined by the following formula:
mol solute
m=
kg solvent
We’ve already assumed that we have 0.100
mol of solute, and we can determine the kg of
H2 O in the usual
way: 18.0 g
1 kg
0.900 mol H2 O
1 mol
1000 g
= 0.0162 kg,
and we can calculate the molality of this solution:
0.100 mol
m=
= 6.17 m
0.0162 kg H2 O
So, it is possible to determine the molality of
this solution without any additional information.
Msci 14 0818
017 10.0 points
The molecular weight of sugar is 342 and that
of water is 18.01. What is the mole fraction
of sugar in a 2.00 molal solution of sugar
dissolved in water?
1. mole fraction = 0.0348 correct
2. mole fraction = 0.950
3. mole fraction = 0.406
4. mole fraction = 0.0360
5. mole fraction = 0.925
Explanation:
MWsugar = 342
msugar = 2.00 m
MWwater = 18.01
mol solute
kg solvent
2.00 mol sugar
Molality of sugar =
1 kg water
Mol water in 1 kg water
1 mol water = (1000 g water)
18.01 g water
= 55.5 mol H2 O
Molality =
Version 001 – Calculating Concentrations WKST – vanden bout – (51165)
6
The mole fraction of sugar is
Xsugar
mol sugar
=
mol sugar + mol water
2.00 mol
=
2.00 mol + 55.5 mol
= 3.48 × 10−2
Nsci 14 0802exam
018 10.0 points
11.6 g of NaCl (58.4 g/mol) and 16.9 g of KBr
(119.8 g/mol) were dissolved in 49 g of water
(18.0 g/mol). What is the mole fraction of
KBr in the solution?
Correct answer: 0.0312.
Explanation:
mNaCl = 11.6 g
mKBr = 16.9 g
mwater = 49 g
FWNaCl = 58.4 g/mol
FWKBr = 119.8 g/mol
FWwater = 18.0 g/mol
11.6 g
= 0.096828 mol
119.8 g/mol
16.9 g
=
= 0.289384 mol
58.4 g/mol
49 g
= 2.72222 mol
=
18.0 g/mol
nKBr =
nNaCl
nH 2 O
Since ntotal = nKBr + nNaCl + nH2 O ,
ntotal = 0.096828 mol + 0.289384 mol
+ 2.72222 mol
= 3.10843 mol ,
the mole fraction is
nKBr
ntotal
0.096828 mol
3.10843 mol
= 0.0312 mol
XKBr =
Nsci 14 0805
019 10.0 points
How many grams of sucrose (C12 H22 O11 )
must be dissolved in 100 g water to make
a solution in which the mole fraction of sucrose is 0.3?
Correct answer: 814.286 g.
Explanation:
mwater = 100 g
nwater =
Xsucrose =
Xsucrose = 0.3
100 g
= 5.55556 mol
18.0 g/mol
n
ntotal
n
0.3 =
n + nwater
n = 0.3 (n + nwater )
0.7 n = 0.3 nwater
0.3 nwater
m
n=
=
, so
0.7
342 g/mol
(342 g/mol) (0.3) (5.55556 mol)
m=
0.7
= 814.286 g
Nsci 14 0807
020 10.0 points
A solution is made from 596 mL methanol
(CH3 OH of density 0.800 g/mL) and 82 mL
of water (H2 O of density 1.000 g/mL). Assume that the solution behaves ideally and
the volumes are additive. Calculate the mole
fraction of methanol in this solution.
Correct answer: 0.765848.
Explanation:
Vmethanol = 596 mL
Vwater = 82 mL
densitymethanol = 0.800 g/mL
densitywater = 1.00 g/mL
mole fraction CH3 OH =
nCH3 OH
ntotal
nCH3 OH = 82 mL CH3 OH
0.800 g CH3 OH
×
1.0 mL CH3 OH
1.0 mL CH3 OH
×
32.0 g CH3 OH
= 14.9 mol CH3 OH
Version 001 – Calculating Concentrations WKST – vanden bout – (51165)
7
1.0 mL H O ntotal = 0.14 mol
Ptotal = 700 torr
2
nH2 O = 82 mL H2 O
nH2 = 0.02 mol
18.0 g H2 O
= 4.55556 mol H2 O
The mole fraction is
XH2 =
14.9 mol
14.9 mol + 4.55556 mol
= 0.765848
XCH3 OH =
Holt da 10 rev 39
021 10.0 points
Three of the primary components of air are
carbon dioxide, nitrogen, and oxygen. In a
sample containing a mixture of only these
gases at exactly one atmosphere pressure, the
partial pressures of carbon dioxide and nitrogen are given as PCO2 = 0.285 torr and
PN2 = 578.351 torr. What is the partial pressure of oxygen?
Correct answer: 181.364 torr.
Explanation:
PCO2 = 0.285 torr
PN2 = 578.351 torr
PT
PO 2
PO 2
PO 2
PT = 1 atm = 760 torr
PO 2 = ?
= PCO2 + PN2 + PO2
= PT − (PCO2 + PN2 )
= 760 torr − (0.285 torr + 578.351 torr)
= 181.364 torr
Mlib 04 1003
022 10.0 points
A 22.4 L vessel contains 0.02 mol H2 gas,
0.02 mol N gas, and 0.1 mol NH3 gas. The
total pressure is 700 torr. What is the partial
pressure of the H2 gas?
1. 100 torr correct
2. 7 torr
3. None of these
4. 28 torr
5. 14 torr
Explanation:
nH 2
0.02 mol
= 0.142857
=
ntotal
0.14 mol
PH2 = XH2 Ptotal
= (0.142857) (700 torr)
= 100 torr