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(E) 0.1kg of water vapor at 110
°C
The correct answer is (B). This type of question is usually solved in the following manner:
(1) .1kg of ice at –20
°C → .1kg of ice at 0°C (melting temperature of ice). This requires .1 kg x
.5kcal/kg
°C × 20°C = 1kcal and leaves 14kcal of unused heat.
(2) .1kg of ice at 0
°C → .1kg of water at 0°C (remember there is no temperature change during a
change of state). This requires .1kg 
× 80kcal/kg = 8kcal and leaves 6kcal.
(3) .1kg of water at 0
°C → .1kg of water at ?°C
6kcal/(.1kg x 1kcal/kg
°C) = 60°C.
There is enough heat to raise the temperature of the water to 60
°C.
EXAMPLE:
A compound is 16% aluminum, 28% sulfur, and 56% oxygen. If the molecular weight is 342,
then the molecular formula is (Atomic weights: Al = 27; S = 32; O = 16)
(A) Al
2
(SO
4
) 
2
(B) Al(SO
4
)
3
(C) Al
3
(SO
4
) 
2
(D) Al
2
(SO
4
)
3
(E) Al
2
S
3
O
8

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Chemistry Subject Test
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The correct answer is (D). The easiest way to solve this problem is to assume that there is 100g of the
unknown compound:
100g of compound
= 16.0g of Al/27g/mole = .6 moles of Al
= 28.0g of S/32g/mole = .875 moles of S
= 56.0g of O/16g/mole = 3.5 moles of O
therefore the empirical formula = Al
.6
S
.875
O
3.5
(divide the subscripts by .6) or AlS
1.5
O
6
.
Empirical weight = (1 
× 27) + (1.5 × 32) + (6 × 16) = 171
Molecular weight = 342
So, the molecular formula = Al
2
S
3
O
12
or Al
2
(SO
4
)
3
.