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dalvyx [7]
3 years ago
7

The nonvolatile, nonelectrolyte testosterone, C19H28O2 (288.40 g/mol), is soluble in benzene C6H6. Calculate the osmotic pressur

e generated when 12.4 grams of testosterone are dissolved in 168 ml of a benzene solution at 298 K.

Chemistry
2 answers:
egoroff_w [7]3 years ago
7 0

Answer: 6.26atm

Explanation:Please see attachment for explanation

BaLLatris [955]3 years ago
3 0

Answer:

The osmotic pressure is 6.26 atm

Explanation:

Step 1: Data given

Mass of testosterone = 12.4 grams

Volume of benzene = 168 mL

Temperature = 298 Kelvin

Step 2: Calculate moles of testosterone

Moles testosterone = mass / molar mass

Moles testosterone = 12.4 grams / 288.42 g/mol

Moles testosterone = 0.0430 moles

Step 3: Calculate molarity

Molarity = moles / volume

Molarity = 0.0430 moles / 0.168 L

Molarity = 0.256 M

Step 4 : Calculate the osmotic pressure

π = iMRT

⇒ with i = The Van't hoff factor = 1 (since testosterone is nonelectrolyte)

⇒ with M = the molair concentration = 0.256 M

⇒ with R = the gas constant = 0.08206 L*atm/k*mol

⇒ with T = the temperature = 298 K

π = 1*0.256*0.08206*298

π = 6.26 atm

The osmotic pressure is 6.26 atm

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6 0
3 years ago
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katen-ka-za [31]

Question: Baking a Cake Without Flour.

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3 years ago
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alekssr [168]

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3 0
3 years ago
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The temperature in a town started out at 55 degrees. By the end of the day the temperature dropped to -6 degrees.
garik1379 [7]

Answer:

The choice of the answer is fourth option that is -61 degrees.

Therefore the temperature drop is -61°Centigrade.

Explanation:

Given:

The temperature in a town started out at 55 degrees

Start temperature = 55°Centigrade. (Initial temperature)

End of the Day      = -6°Centigrade. (Final temperature)

To Find:

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We will have,

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Substituting the above values in it we get

\textrm{Temperature drop}=-6-55\\\\\therefore \textrm{Temperature drop}=-61\° centigrade

Therefore the temperature drop is -61°Centigrade.

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