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Soloha48 [4]
3 years ago
14

A sample of an unknown metal has a mass of 6.557 g. The metal was carefully added to a graduated cylinder containing 10.50 mL of

water. The water level in the graduated cylinder rose to 11.16 mL. What is the density of the unknown metal?​
Chemistry
1 answer:
Levart [38]3 years ago
8 0

Answer:

Explanation:

Be careful. The tricky part of the problem is that there are 4 places of sig digs.

m = 6.557 grams

V = 11.16 - 10.50 = 0.66

density = mass/ volume

density = 6.557/0.66 = 9.935 g/mL

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Total vapor pressure can be calculated using partial vapor pressures and mole fraction as follows:

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Here, X_{A} is mole fraction of A, X_{B} is mole fraction of B, P_{A} is partial pressure of A and P_{B} is partial pressure of B.

The mole fraction of A and B are related to each other as follows:

X_{A}+X_{B}=1

In this problem, A is hexane and B is octane, mole fraction of hexane is given 0.580 thus, mole fraction of octane can be calculated as follows:

X_{octane}=1-X_{hexane}=1-0.58=0.42

Partial pressure of hexane and octane is given 183 mmHg and 59.2 mmHg respectively.

Now, vapor pressure can be calculated as follows:

P=X_{hexane}P_{hexane}+X_{octane}P_{octane}

Putting the values,

P=(0.580)(183 mmHg)+(0.420)(59.2 mmHg)=131 mmHg

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