1
CHM2046 TEST 1 Version A (Exam 3)
Answers underlined
N
A
= 6.022×10
23
;
R = 8.314 J/mol.K = 0.08206 L.atm/mol.K; 760 torr = 760 mmHg = 1 atm
1)
A 150.0 mL sample of an aqueous solution at 25°C contains 15.2
mg of an unknown nonelectrolyte compound.
If the solution has
an osmotic pressure of 8.44 torr, what is the molar mass of the
unknown compound?
a)
195 g/mol
b)
223 g/mol
c)
294 g/mol
d)
341 g/mol
e)
448 g/mol
Use
Π
=MRT; solve for M; solve for molar mass
M =
Π
/RT = (8.44/760)/ (0.08206
×
298) = 4.541
×
10
-4
mol/L
mol = (4.541
×
10
-4
mol/L)(0.150 L)= 6.812
-5
mol
molar mass = 0.0152 g/6.812
-5
mol = 223g/mol
2)
Determine the vapor pressure of a solution at 25°C that contains
76.6 g of glucose (C
6
H
12
O
6
) in 250.0 mL of water.
The vapor
pressure of pure water at 25°C is 23.8 torr.
a)
23.1 torr
b)
22.9 torr
c)
70.8 torr
d)
72.9 torr
e)
7.29 torr
Use P
solvent
=
χ
solvent
P
o
solvent
= 0.970
×
23.8 = 23.1torr
3)
Determine
∆
H
vap
for a compound that has a measured vapor
pressure of 24.3 torr at 273 K and 135 torr at 325 K.
Use Clausius-Clapeyron eqn; solve for
∆
H
vap
⎟
⎠
⎞
⎜
⎝
⎛
−
∆
=
325
1
273
1
3
.
24
135
ln
R
H
vap
;
1.71 =
∆
H
vap/R (5.83
×
10
-4
)
∆
H
vap = (1.71
×
8.314)/(5.83
×
10
-4
)
÷
1000 J

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