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Delicious77 [7]
3 years ago
11

A solution of pentane and ethanol (CH3CH2OH)that is 50.% pentane by mass is boiling at 57.2°C. The vapor is collected and cooled

until it condenses to form a new solution.
Calculate the percent by mass of pentane in the new solution. Here's some data you may need:

normal boiling point density vapor pressure at
57.2°C
pentane 36.°C 0.63gmL 1439.torr
ethanol 78.°C 0.79gmL 326.torr
Be sure your answer has 2 significant digits.
dont round during math only for answer!
Note for advanced students: you may assume the solution and vapor above it are ideal.
Chemistry
1 answer:
Paladinen [302]3 years ago
6 0

Answer:

The correct answer is 81.52 percent.

Explanation:

Based on the given information, the boiling point of pentane is 36 degree C and the boiling point of ethanol is 78 degree C. The density of pentane and ethanol is 0.63 g/ml and 0.79 g/ml. The vapor pressure of pentane at 57.2 degree C is 1439 torr and the vapor pressure of ethanol at 57.2 degree C is 326 torr.  

In the given case, 50 percent pentane by mass signifies that mass of pentane is 50 grams. Thus, the mass of ethanol will be 100-50 = 50 grams.  

The moles or n can be calculated by using the formula,  

n = weight/molecular mass

The molecular mass of pentane is 72.15 g per mol and the molar mass of ethanol is 46.07 g/mol.  

The moles of pentane is,  

= 50 g/72.15 g/mol = 0.6930 mol

The moles of ethanol is,  

= 50 g/46.07 g/mol = 1.0853 mol

The mole fraction of pentane is,  

= 0.6930 mol / (0.6930 + 1.0853) mol = 0.3897  

The mole fraction of ethanol is,  

= 1.0853 mol / (0.6930 + 1.0853) mol = 0.6103

Now the vapor pressure of solution will be,  

= pressure of pentane * mole fraction of pentane + pressure of ethanol * mole fraction of ethanol

= (1439 * 0.3897) + (326 * 0.6103)

= 759.736 torr

The vapor pressure of pentane within the solution,  

= vapor pressure of pentane * mole fraction of pentane

= 1439 torr * 0.3897

= 560.778 torr

The fraction of pentane is,  

= 560.778 / 759.736 = 0.738

Let us assume that the total mole is 1, the mole fraction of pentane is 0.738, so the mole fraction of ethanol will become, 1-0.738 = 0.262

The mass of pentane = 0.738 * 72.15 = 53.2467

The mass of ethanol = 0.262 * 46.07 = 12.07034

The percent by mass of pentane in new solution will be,  

Mass% = mass of pentane/Total mass * 100%

= 53.2467/(53.2467 + 12.07034) * 100%

= 53.2467/65.31704 * 100 %

= 81.52 %

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<h3>Answer:</h3>

62.3 mol Ca

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
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7 0
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A 73.6 g sample of aluminum is heated to 95.0°C and dropped into 100.0 g of water at 20.0°C. If the resulting temperature of the
Softa [21]

Answer:

The specific heat of aluminium is 0.875 J/g°C

Explanation:

Step 1: Data given

The mass of the aluminium sample = 73.6 grams

Initial temperature of the sample = 95.0 °C

Mass of water = 100.0 grams

Initial temperature of water = 20.0 °C

Final temperature of water and aluminium = 30.0 °C

The specific heat of water = 4.184 J/g°C

Step 2: Calculate the specific heat of aluminium

Q gained = Q lost

Qwater = -Qaluminium

Q =  m*c*ΔT

m(aluminium) * c(aluminium) * ΔT(aluminium) = - m(water) * c(water) * ΔT(aluminium)

⇒ mass of aluminium = 73.6 grams

⇒ c(aluminium) = TO BE DETERMINED

⇒ ΔT(aluminium) = The change of temperature = T2 - T1 = 30 .0 °C - 95.0 °C = -65.0°C

⇒ mass of water = 100.0 grams

⇒ c(water ) = The specific heat of water = 4.184 J/g°C

⇒ ΔT(water) = The change of temperature of water = T2 - T1 = 30.0 - 20.0 = 10.0 °C

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c(aluminium) = 0.875 J /g°C

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