Chem 1412 homework chap12
1.
Which of the following has very high solubility in water?
A)
C6H6
B)
C2H5OH
C)
C6H5NH2
D)
C6H5OH
Hint: like dissolve like. Water is a polar molecule. So polar molecules have very high solubility in water.
The molecules having a benzene ring usually are not soluble in polar solution, like water.
3.
Which of the following is more soluble in benzene than in water?
A)
potassium chloride
B)
naphthalene
C)
washing soda
D)
CsF
Hint: benzene is a non-polar solvent. Like disovle like
4.
12 g of urea (molar mass = 60 g) is dissolved in 180 g of water. The mole fraction of urea is :
A)
0.20
B)
0.066
C)
0.020
D)
0.66
Hint:X1=n1/(n1+n2); n1=12g/(60g/mol)=0.2mol; n2=180g/(18g/mol)=10mol,X1=0.2/(0.2+10)mol=0.02
5.
6.00 g of urea (molar mass = 60 g) is dissolved in 100 g of water (M.W – 18). The percent by
mass of urea in the solution is _____.
A)
5.7%
B)
6.0%
C)
16.6%
D)
3.0%
Hint: mass %=6.00g/(6.00g+100g)=5.7%
6.
The molality of 14.3 g of sucrose (C12H22O11) in 676 g of water is _____.
A)
0.0210 m
B)
2.03 m
C)
0.0619 m
D)
1.09 m
Hint:m=n/mass of solvent(kg), n=14.3g/342(g/mol)=0.048mol, m=0.048mol/0.676kg=0.0691m
7.
What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is 0.927 g/mL?
A)
1.53 m
B)
0.68 m
C)
1.68 m
D)
8.92 m
m =
moles of solute
M =
mass of solvent
moles o
liters of solution
Assume 1 L of solution:
5.86 moles ethanol = 270 g eth
= 927 g – 927
270 gg of
= 657
g
= 0.657
solution
(1000
mL x 0.
mass
of solvent
mass of solvent = g/mL)
mass of moles
solution
mass
of –solute
m =
of solute
8.
A)
B)
C)
D)
66.
A)
B)
C)
D)
The molality of a 48.2% by mass of KBr solution is _____.
3.42 m
7.82 m
5.12 m
10.08 m
The boiling point of a 2.47 m naphthalene in benzene solution (b.p 80.1◦C ) is: (Kb=2.53◦C/m)
86.3 ◦C
82.1 ◦C
83.7 ◦C
84.5 ◦C


Hint: ΔTb = Kb m =(2.53◦C/m)x2.47m=6.25◦C, ΔTb = Tb – T b◦ , Tb= ΔTb +T b◦=6.25+80.1=86.3
9.
An aqueous solution of a nonelectrolyte freezes at –1.1 ◦C. The molality of the solution is
(Kf=1.86◦C/m).
A)
0.49 m
B)
0.035 m
C)
0.59 m
D)
1.26 m
Hint: ΔTf = Kf m,m= ΔTf / Kf=1.1/1.86=0.59
10.
The freezing point of 21.2 g of NaCl in 135 mL of water(Kf=1.86◦C/m) is _____.
A)
–5.00 ◦C
B)
–4.60 ◦C
C)
–9.99 ◦C
D)
–11.5 ◦C
Hint: ΔTf = iKf m, m= n of solute/mass of solvent(kg)=[21.2g/(58.45g/mol)]/0.135kg=2.69m,
ΔTf = iKf m=2x(1.86◦C/m )x2.69m=9.99◦C, ΔTf = T f◦-Tf , Tf=0-9.99=-9.99◦C