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timofeeve [1]
3 years ago
12

g If a sample of 50 mL of ethyl benzene (bp = 136 C) was contaminated with 10 mL of ethyl acetate (bp = 77 C) what effect would

this have on the boiling point measurement?
Chemistry
1 answer:
PolarNik [594]3 years ago
6 0

Answer:

melting point will be higher than that of pure ethyl acetate

Explanation:

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When 25.0 ml of 0.500 m h2so4 is added to 25.0 ml of 1.00 m koh in a coffee-cup calorimeter at 23.50°c, the temperature rises to
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The chemical reaction that occurs between the given substances is a neutralization reaction as shown below :

 

H_2SO_4 + 2KOH -> K_2SO_4 + 2H_2O

<span>1 mol            2 mol      1 mol           2 mol </span>

 

<span>The number of moles of the given substances  is calculated as shown : </span>

Number of moles of H_2SO_4 = 25.0 mL x 0.50 M = 12.5 millimoles

Number of moles of KOH = 25.0 mL x 1.00 M = 25.0 millimoles

 

As 1 mol of sulfuric acid reacts with 2 mol of KOH to give 2 mol of water, 12.5 millimoles of sulfuric acid completely reacts with 25.0 millimoles of KOH to give 25.0 millimoles of water.

Total volume of the solution = 25.0 mL + 25.0 mL = 50.0 mL.

Density of water is 1 g/mL. Use this to calculate the mass of the solution.

<span>Mass of the solution – 50.0 mL x 1 g/mL = 50.0  </span>

The specific heat of water is 4.184 J/gK. The temperature of the solution is increased from 23.5 degrees Celsius to 30.17 degrees Celsius.

The amount of heat released = 4.184 J/gK x 50.0 g x (30.17C – 23.50C) 1395 J

 

The amount of heat released per one mole of water formed can be calculated as shown :

The amount of heat released for formation of mole water = 1395 J / (25.0 m mol x 1mol/1000 m mol)

= 55,800 J

<span> </span>

8 0
3 years ago
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