Practice Problems for Test 2
Which one of the following causes an increase in the solubility of a gas in a given solvent in the situation in which the gas does not react with the solvent to form a new substance? a. increasing the temperature of the solvent and simultaneously decreasing the partial pressure of the gas in the space above the solvent b. decreasing the temperature of the solvent and simultaneously increasing the partial pressure of the gas in the space above the solvent c. increasing the temperature of the solvent and simultaneously increasing the partial pressure of the gas in the space above the solvent d. decreasing the temperature of the solvent and simultaneously decreasing the partial pressure of the gas in the space above the solvent e. increasing the temperature of the solvent while maintaining the partial pressure of the gas in the space above the solvent at a set value
Concerning the solvation step of the solution process, the process is a. exothermic and the potential energy increases b. exothermic and the potential energy decreases c. endothermic and the potential energy increases d. endothermic and the potential energy decreases e. endothermic and occurs with no change in potential energy
The solubility of O
2
in water is approximately 0.00380 grams per liter of water when the temperature is 25.0 o C and the partial pressure of gaseous oxygen is 760 torr. What should the solubility of oxygen be if the oxygen pressure is readjusted to 1000 torr? a. 0.00289 g L -1 b. 0.00500 g L
-1 c. 1.49 g L
-1 d. 2.89 x 10 3 g L -1 e. 3.46 x 10
3
g L
-1
A solution of potassium nitrate is prepared by mixing 3.50 g of KNO
3
with 12.0 g of water. The percent, by mass, of KNO
3
is a. 22.6 % b. 23.3 % c. 28.0 % d. 29.2 % e. 41.8 %

How many grams of NaC
2
H
3
O
2
(M = 82.034 g mol -1 ) should be dissolved in 400.0 g of water to prepare a solution which is 11.28 % NaC
2
H
3
O
2
by mass? a. 3.146 g b. 7.558 g c. 21.17 g d. 50.86 g e. 127.15 g
A solution is prepared by mixing 0.3355 moles of NaNO
3
(84.995 g mol
-1
) with 235.0 g of water (18.015 g mol -1 ). Its density is 1.0733 g mL -1 . The solute is completely dissociated. What is the molality of the solution? a. 0.6474 m b. 0.7004 m c. 1.320 m d. 1.428 m e. 1.545 m
At 28.0 o
C, the vapor pressure of acetonitrile, CH
3
CN, is 102.0 torr while that of acetone,
C
3
H
6
O, is 228.9 torr and for CS
2
the value is 378.7 torr. A three component solution is made by adding 0.300 moles of CH
3
CN and 0.400 moles of C
3
H
6
O to 0.350 moles of CS
2
. The mixture behaves as an ideal mixture. What is the vapor pressure of the solution? a. 203 torr b. 243 torr c. 262 torr d. 275 torr e. 610 torr
Pure cyclohexane, C
6
H
12
, has a freezing point of 6.53 o
C. Its freezing point depression constant is: K f
= 20.0 o
C m
-1
. A solution was made by taking 11.40 g of an unknown non-electrolyte and dissolving it in 150.0 g of cyclohexane. The measured freezing point of the solution was -0.78 o
C. Calculate the molecular weight of the unknown substance. a. 27.8 g mol -1 b. 46.8 g mol
-1 c. 208 g mol
-1 d. 264 g mol -1 e. 1949 g mol
-1

The reaction, 2 NO( g ) + O
2
( g )

2 NO
2
( g ), was found to be first order in each of the two reactants and second order overall. The rate law should therefore be written as a. rate = k[NO] 2 b. rate = k([NO][O
2
] c. rate = k[NO
2
]
2
[NO]
-2
[O
2
]
-½ d. rate = k[NO] 2 [O
2
] 2 e. rate = k([NO][O
2
])
2
For the reaction, A + 2 B

C + 2 D, some measurements of the initial rate of reaction at varying concentration gave the following data.
run # [A] [B] rate, mol L
-1
s
-1
1 0.100 0.200 0.000360
2 0.150 0.200 0.000540
3 0.150 0.250 0.001055 a. the rate law is therefore: rate = k[A]
2
[B] b. the rate law is therefore: rate = k[A][B]
2 c. the rate law is therefore: rate = k[A]
2
[B]
2 d. the rate law is therefore: rate = k[A][B] e. the rate law is therefore: rate = k[A][B] 3
The rate constant for a first order decomposition reaction is 0.0111 min
-1
. What is the half-life of the reaction? a. 111 min b. 62.4 min c. 5000 sec d. 31.25 min e. 27.1 min
In a first order reaction, what fraction of the material will remain after 4 half-lives? a. 1/16 b. 1/8 c. 1/9 d. 1/4
e. 1/3

For a one step reaction, the activation energy for the forward reaction is 40.0 kJ mol
-1
, and the enthalpy of reaction is -20.0 kJ mol
-1
. Calculate the activation energy for the reverse reaction. a. +60.0 kJ mol
-1 b. -20.0 kJ mol -1 c. -1200 kJ kJ mol
-1 d. +20.0 kJ kJ mol
-1 e. +1200 kJ kJ mol
-1
The activation energy for a reaction can be found by finding the slope of a plot of ln (k) vs
T
-1
and a. adding this slope to -R b. multiplying this slope by 2.303 c. dividing this slope by -R d. multiplying this slope by 2.303R e. multiplying this slope by -R