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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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The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: OCl−+I−→OI−+Cl− T
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Answer :

(a) The rate law for the reaction is:

\text{Rate}=k[OCl^-]^1[I^-]^1

(b) The value of rate constant is, 60.4M^{-1}s^{-1}

(c) rate of the reaction is 6.52\times 10^{-5}Ms^{-1}

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

OCl^-+I^-\rightarrow OI^-+Cl^-

Rate law expression for the reaction:

\text{Rate}=k[OCl^-]^a[I^-]^b

where,

a = order with respect to OCl^-

b = order with respect to I^-

Expression for rate law for first observation:

1.36\times 10^{-4}=k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b ....(1)

Expression for rate law for second observation:

2.72\times 10^{-4}=k(3.0\times 10^{-3})^a(1.5\times 10^{-3})^b ....(2)

Expression for rate law for third observation:

2.72\times 10^{-4}=k(1.5\times 10^{-3})^a(3.0\times 10^{-3})^b ....(3)

Dividing 1 from 2, we get:

\frac{2.72\times 10^{-4}}{1.36\times 10^{-4}}=\frac{k(3.0\times 10^{-3})^a(1.5\times 10^{-3})^b}{k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b}\\\\2=2^a\\a=1

Dividing 1 from 3, we get:

\frac{2.72\times 10^{-4}}{1.36\times 10^{-4}}=\frac{k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b}{k(1.5\times 10^{-3})^a(3.0\times 10^{-3})^b}\\\\2=2^b\\b=1

Thus, the rate law becomes:

\text{Rate}=k[OCl^-]^a[I^-]^b

a  = 1 and b = 1

\text{Rate}=k[OCl^-]^1[I^-]^1

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1.36\times 10^{-4}=k(1.5\times 10^{-3})(1.5\times 10^{-3})

k=60.4M^{-1}s^{-1}

Now we have to calculate the rate for a reaction when concentration of OCl^-  and I^-  is 1.8\times 10^{-3}M and 6.0\times 10^{-4}M respectively.

\text{Rate}=k[OCl^-][I^-]

\text{Rate}=(60.4M^{-1}s^{-1})\times (1.8\times 10^{-3}M)(6.0\times 10^{-4}M)

\text{Rate}=6.52\times 10^{-5}Ms^{-1}

Therefore, the rate of the reaction is 6.52\times 10^{-5}Ms^{-1}

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