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Mariana [72]
3 years ago
13

E ) The distribution coefficient , Ko ( Cether / C water ) , for an organic substance X at room temperature is 13. What relative

volume of ether to water should be used for the extraction of 94 percent of X from a water solution in single extraction ?
Chemistry
1 answer:
TiliK225 [7]3 years ago
4 0

Answer:

Relative volume of ether to water that should be used for the extraction = 1.205

Explanation:

The extraction/distribution coefficient of an arbitrary solvent to water for a given substance is expressed as the mass concentration of the substance in the arbitrary solvent (C₁) divided by the mass concentration of the substance in water (C₂).

K = (C₁/C₂)

Let the initial mass of the organic substance X in water be 1 g (it could be any mass basically, it is just to select a right basis, since we are basically working with percentages here).

If 94% of the organic substance X is extracted by ether in a single extraction, 0.94 g ends up in ether and 0.06 g of the organic substance X that remains in water.

Let the volume of ether required be x mL.

Let the volume of water required be y mL.

Relative volume of ether to water that should be used for the extraction = (x/y)

Mass concentration of the organic substance X in ether = (0.94/x)

Mass concentration of organic substance X in water = (0.06/y)

The distribution coefficient , Ko (Cether / C water), for an organic substance X at room temperature is 13.

13 = (0.94/x) ÷ (0.06/y)

13 = (0.94/x) × (y/0.06)

13 = (15.667y/x)

(x/y) = (15.667/13) = 1.205

Hope this Helps!!!

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cestrela7 [59]
In 1 mol of CH3OH, you have 4 H-atoms (because 3 H-atoms are attached to the C-atom, and one H-atom in the OH group). That means in 0.500 mol of CH3OH, you have 2 H-atoms since it is halved. And then we have Avogadro's constant: 6.02 * 1023.

The question asks for how many hydrogen atoms there are in 0.500 mol CH3OH. Using the numbers that we have (Avogadro's constant and no. of H-atoms), the answer of the question will be something like:

<span>H-atoms in CH3OH = 2 * 6.02 * </span>1023<span> = ~1.2 * 10</span>24

 


8 0
3 years ago
If I had 3.50 x 10 24molecules of Cl2 gas, how many grams is this?
Zinaida [17]

Answer:

412 g Cl₂

General Formulas and Concepts:

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Moles
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

[Given] 3.50 × 10²⁴ molecules Cl₂

[Solve] grams Cl₂

<u>Step 2: Identify Conversions</u>

Avogadro's Number

[PT] Molar Mass of Cl - 35.45 g/mol

Molar Mass of Cl₂ - 2(35.45) = 70.9 g/mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 3.50 \cdot 10^{24} \ molecules \ Cl_2(\frac{1 \ mol \ Cl_2}{6.022 \cdot 10^{23} \ molecules \ Cl_2})(\frac{70.9 \ g \ Cl_2}{1 \ mol \ Cl_2})
  2. [DA] Divide/Multiply [Cancel out units]:                                                           \displaystyle 412.072 \ g \ Cl_2

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

412.072 g Cl₂ ≈ 412 g Cl₂

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