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ahrayia [7]
3 years ago
12

Imagine two solutions with the same concentration and the same boiling point, but one has benzene as the solvent and the other h

as carbon tetrachloride as the solvent. Determine that concentration and boiling point.
Chemistry
2 answers:
natima [27]3 years ago
5 0
The Boiling point of the benzene would be

bp= 80.1C
 kb=2.53

The boiling point of the carbon tetrachrloride would be
bp=80.1
 kb=50.3 
salantis [7]3 years ago
5 0

Answer:

The concentration should be 1.32 M and the boiling point 83.44 °C

Explanation:

We have two solutions with the same concentration and the same boiling point.

The first solution is benzene as the solvent and the other has carbon tetrachloride as the solvent.

Useful data on the problem:

Boiling point Benzene pure is 80.1 °C and Kbb = 2.53

Boling point carbon tetrachloride pure 76.8 °C and Kbc = 5.03.

The boiling point of the first solution should be:

B1 = 80.1 + 2.53*X1. Where X1 is the concentration in M (mol/L)

The boiling point of the second solution is:

B2 = 76.8 + 5.03*X2. Where X2 is the concentration in M (mol/L).

Both solutions have the same concentration and the same boiling point. We can use the equalities X1 = X2 = X. B1 = B2

B1 = B2

80.1 + 2.53*X = 76.8 + 5.03*X

We can isolate X.

80.1 - 76.8 = (5.03 - 2.53)X

3.3 = 2.5 X

2.5 X = 3.3

X = 3.3/2.5

X = 1.32 M

The concentration of both solutions X1 = X2 = 1.23 M and the boiling point is

80.1 + 2.53(1.32) = 83.44 °C.

Finally, we can conclude that the concentration should be 1.32 M and the boiling point 83.44 °C.

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How many moles of Carbon are in 3.06 g of Carbon
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Answer:

\boxed {\boxed {\sf 0.255 \ mol \ C }}

Explanation:

If we want to convert from grams to moles, the molar mass is used. This is the mass of 1 mole. They are found on the Periodic Table as the atomic masses, but the units are grams per mole (g/mol) instead of atomic mass units (amu).

Look up the molar mass of carbon.

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Set up a ratio using the molar mass.

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Since we are converting 3.06 grams to moles, we multiply by that value.

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Flip the ratio. This way, the ratio is still equivalent, but the units of grams of carbon cancel.

3.06 \ g \ C* \frac{1 \ mol \ C}{12.011 \ g\ C}                      

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\frac {3.06}{12.011 } \ mol \ C                                

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The original measurement of grams (3.06) has 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandth place.

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The 7 in the ten-thousandth place tells us to round the 4 up to a 5.

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