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Anon25 [30]
3 years ago
9

A 5.00g of X, the product of organic synthesis is obtained in a 1.0 dm3 aqueous solution. Calculate the mass of X that can be ex

tracted from the aqueous solution by a 50cm3 of ethoxy ethane. (KD (X) =40.​
Chemistry
1 answer:
Anestetic [448]3 years ago
8 0

Answer:

mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g

Explanation:

The partition coefficient of X between ethoxy ethane (ether) and water, K is given by the formula

K = concentration of X in ether/concentration of X in water

Partition coefficient, K(X) between ethoxy ethane and water = 40

Concentration of X in ether = mass(g)/volume(dm³)

Mass of X in ether = m g

Volume of ether = 50/1000 dm³ = 0.05 dm³

Concentration of X in ether = (m/0.05) g/dm³

Concentration of X in water = mass(g)/volume(dm³)

Mass of X in water left after extraction with ether = (5 - m) g

Volume of water = 1 dm³

Concentration of X in water = (5 - m/1) g/dm³

Using K = concentration of X in ether/concentration of X in water;

40 = (m/0.05)/(5 - m)

(m/0.05) = 40 × (5 - m)

(m/0.05) = 200 - 40m

m = 0.05 × (200 - 40m)

m = 10 - 2m

3m = 10

m = 10/3

m = 3.33 g of X

Therefore, mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g

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What is the balanced, net ionic equation
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<h3>What is a reaction equation?</h3>

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In this case, the complete molecular equation can be written as;

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3 0
2 years ago
A chemist prepares a solution of potassium permanganate by measuring out 26. g of potassium permanganate into a 350. mL volumetr
alekssr [168]

Answer:

0.471 mol/L

Explanation:

First, we'll begin by by calculating the number of mole of KMnO4 in 26g of KMnO4.

This is illustrated below:

Molar Mass of KMnO4 = 39 + 55 + (16x4) = 39 + 55 + 64 = 158g/mol

Mass of KMnO4 from the question = 26g

Mole of KMnO4 =?

Number of mole = Mass/Molar Mass

Mole of KMnO4 = 26/158 = 0.165mole

Now we can obtain the concentration of KMnO4 in mol/L as follow:

Volume of the solution = 350mL = 350/1000 = 0.35L

Mole of KMnO4 = 0.165mole

Conc. In mol/L = mole of solute(KMnO4)/volume of solution

Conc. In mol/L = 0.165mol/0.35

conc. in mol/L = 0.471mol/L

3 0
3 years ago
Read 2 more answers
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