Chemistry 1412
Chapter 12
Homework
1) Which of the following concentration unit will be affected by temperature?
(a) Mass percent
(b) mole fraction
(c) molarity (d) molality (e) ppm
2) Which of the following action will reduce the solubility of a gas in water?
(a) Pressure (b) temperature (c) molar mass of the gas (d) acidic or basic gas
3) A 0.270 M KOH solution has a density of 1.01 g/mL. Calculate the molality (m)
of KOH. (a) 0.271 (b) 0.273 (c) 0.265
(d) 0.267
4) Which of the following pair is miscible (i.e. soluble to each other)?
(a) CH3OH and C6H6 (b) CCl4 and H2O (c) C6H12O6 and C2H5OH
(d) CH3Cl and CH3COOH
5) What’s the concentration if the osmotic pressure for sucrose, C12H22O11, solution
is 0.048 atm at 20oC?
(A)1.9M
(B)0.25 M (C) 0.002M
(D)2
0.002 M
p = MRT, M=p/RT=0.048 atm/[(0.082atm·L/K·mol)x293K]=0.002M
6) The osmotic pressure of 0.050 M MgSO4 at 25
Van't Hoff factor?
(A)1.9
(B)1.3 (C)2.5
C is 1.60 atm. What is the
(D)2
p = iMRT, i=p/MRT=1.60 atm/[0.050Mx(0.082atm·L/K·mol)x298K]=1.3
7) What mass of sucrose (C12H22O11) in kilograms must be added to 552 g of
water to yield a solution with vapor pressure 2.0 mmHg less than that of pure water
at 20 °C (The vapor pressure of water is 17.5 mmHg at 20 °C)?
A)
0.52 kg B) 0.052 kg C)
1.3 kg
D) 3.1 kg
Hint: P10-P1=ΔP=X2 P10, X2 = ΔP/ P10=2.00mmHg/17.5mmHg=0.11=n2/(n1+n2),
n1=552g/18g/mol=30.67mol, solve for n2=3.8mol, mass of sucrose=
n2xM=3.8molx342g/mol=1.3kg
8) Which of the following has very high solubility in water?
A)C6H6
B) C2H5OH C)C6H5NH2
D)C2H5OH
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.
9) 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
10) 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
11) 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%
12) 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
13) What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is
0.927 g/mL?
A)
B)
C)
D)
1.53 m
0.68 m
1.68 m
8.92 m
14) The molality of a 48.2% by mass of KBr solution is _____.
A)
3.42 m
B)
7.82 m
C)
5.12 m
D)
10.08 m
15) The boiling point of a 2.47 m naphthalene in benzene solution (b.p 80.1◦C ) is:
(Kb=2.53◦C/m)
A)
86.3 ◦C
B)
82.1 ◦C
C)
83.7 ◦C
D)
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
16) 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
17)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
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
18) A solution of 12.00 g of an unknown non dissociating compound dissolved in
200.0 g of benzene freezes at 3.45oC. The Kf of benzene is 5.07oC/m. The freezing
point of benzene is 5.45oC.
(a) What is the mass percent of the solution?
12.00g/(12.00+200.0)g=5.66%
(b) What is the molality of the unknown solution?
ΔTf = Kf m, ΔTf = T f ◦ – Tf = 5.45-3.45=2.00 oC, m= ΔTf /Kf= 2.00 oC/(5.07oC/m)=0.394m
(c) How many moles of unknown in the solution?
The molality m = moles of solute/1kg solvent, in 200.0g of solvent, there are 200.0g x
(1kg/1000g) x 0.394mol/ kg solvent=0.0789 mol
(d) What is the molar mass of the unknown?
M= mass/n=12.00g/0.0789 mol=152.09 g/mol
(e) Assume the density of the unknown solution is 1.35 g/mL, what is the molarity
of the unknown solution?
d= mass/V=1.35 g/ml, For 1L solution, the mass of the solution is (1.35 g/ml)x1000mL= 1350g,
the mass of solute is 5.66% x 1350g= 76.41g, molarity= moles of solute/V= (76.41g/152.09
g/mol)/1L= 0.502 mol/L