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In-s [12.5K]
3 years ago
6

An organic chemist is planning to extract 1.00 g of an organic compound dissolved in 200.0 mL of water into a diethyl ether solv

ent. The partition coefficient for this system is 9.5, favoring the diethyl ether solvent. What percentage of the organic compound remains in the aqueous solvent if you perform three successive extractions with 50.0 mL of diethyl ether for each extraction?(a) 29.63% (b) 2.60% (c) 0.00169%
Chemistry
1 answer:
ddd [48]3 years ago
6 0

Answer:

(b) 2.60%

Explanation:

Partition coefficient of the organic compound in ether is:

9.5 = Concentration in ether / Concentration in water

In the first extraction, X is the amount of organic compound extracted:

9.5 = X/50mL / (1-X)/200mL

9.5 = 200X / (50-50X)

475 - 475X = 200X

475 = 675X

X = 0.7037g are extracted.

Remains = 1g - 0.7037g = 0.2963g

Second extraction:

9.5 = X/50mL / (0.2963-X)/200mL

9.5 = 200X / (14.815-50X)

140.74 - 475X = 200X

140.74 = 675X

X = 0.2085g are extracted.

Remains = 0.2963g - 0.2085g = 0.0878g

Third extraction:

9.5 = X/50mL / (0.0878g-X)/200mL

9.5 = 200X / (4.3899-50X)

41.70 - 475X = 200X

41.70 = 675X

X = 0.2085g are extracted.

Remains = 0.0878g - 0.0618g = 0.026g remains

Percentage: 0.026g / 1g * 100 =

<h3>(b) 2.60%</h3>
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6 0
3 years ago
2. A sample of neon gas has a volume of 87.6 L at STP. How many moles are present?
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Answer:

                      3.91 moles of Neon

Explanation:

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Data Given:

                 n = moles = <u>???</u>

                 V = Volume = 87.6 L

Solution:

               As 22.4 L volume is occupied by one mole of gas then the 16.8 L of this gas will contain....

                          = ( 1 mole × 87.6 L) ÷ 22.4 L

                          = 3.91 moles

<h3>2nd Method:</h3>

                     Assuming that the gas is acting ideally, hence, applying ideal gas equation.

                              P V  =  n R T      ∴  R  =  0.08205 L⋅atm⋅K⁻¹⋅mol⁻¹

Solving for n,

                              n  =  P V / R T

Putting values,

                              n  =  (1 atm × 87.6 L)/(0.08205 L⋅atm⋅K⁻¹⋅mol⁻¹ × 273.15K)

                              n  =  3.91 moles

Result:

          87.6 L of Neon gas will contain 3.91 moles at standard temperature and pressure.

7 0
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