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kiruha
3 months ago
15

Suppose a student needs to standardize a sodium thiosulfate, Na2S2O3,Na2S2O3, solution for a titration experiment. To do so, he

or she will react it with a solution of iodine. The student adds a 1.00 mL1.00 mL aliquot of 0.0200 M KIO30.0200 M KIO3 solution to a flask, followed by 3 mL3 mL of distilled water, 0.2 g0.2 g of solid KI,KI, and 1 mL H2SO4.1 mL H2SO4. The student then titrates the solution with sodium thiosulfate solution in order to determine the exact concentration of Na2S2O3.Na2S2O3. The end point of the titration is reached after 0.90 mL0.90 mL of Na2S2O3Na2S2O3 is dispensed from a microburet. What is the concentration of the standard sodium thiosulfate solution?
Chemistry
1 answer:
alisha [2.9K]3 months ago
5 0

Answer:

0.133

Explanation:

The reaction that occurs between KIO3 and KI in an acidic medium is described as

IO3⁻ +5I⁻ +6h⁺ → 3I₂ + 3H₂O

I₂ subsequently reacts with sodium thiosulfate

NaS₂O₃  → 2Na⁺ + S₂O₃²⁻

The overall reaction can be summarized as

IO⁻₃ + 6H⁺ + 6S₂O₃³⁻ → I⁻ + 3S₄O₆²⁻ + 3H₂O

The mole of KIO₃

is computed using molarity multiplied by volume

\frac{0.02mol}{L} *0.01L

which equals 0.00002mol

One mole of KIO₃ reacts with 6 moles of S₂O₃²⁻

which gives 2x6x10⁻⁵

= 0.00012 mol

The volume is 0.90 ml

1 ml equals 0.001L

0.90ML  is 0.0009L

To find concentration,

molarity/volume

= 0.00012/0.0009

= 0.133m

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"A sample of silicon has an average atomic mass of 28.084amu. In the sample, there are three isotopic forms of silicon. About 92
lorasvet [2795]

Answer: The percentage abundance for _{14}^{30}\textrm{Si} isotope is 3.09 %.

Explanation:

The average atomic mass of an element is calculated by taking the sum of the masses of all isotopes weighted by their respective natural fractional abundances.

To compute average atomic mass, the following formula is applied:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i   .....(1)

From the information provided:

Let the fractional abundance for _{14}^{28}\textrm{Si} isotope be 'x'

  • For _{14}^{28}\textrm{Si} isotope:

Mass of _{14}^{28}\textrm{Si} isotope = 27.9769 amu

Percentage abundance of _{14}^{28}\textrm{Si} isotope = 92.22 %

Fractional abundance for _{14}^{28}\textrm{Si} isotope = 0.9222

  • For _{14}^{29}\textrm{Si} isotope:

Mass of _{14}^{28}\textrm{Si} isotope = 28.9764 amu

Percentage abundance for _{14}^{28}\textrm{Si} isotope = 4.68%

Fractional abundance of _{14}^{28}\textrm{Si} isotope = 0.0468

  • For _{14}^{30}\textrm{Si} isotope:

Mass of _{14}^{30}\textrm{Si} isotope = 29.9737 amu

Fractional abundance for _{14}^{30}\textrm{Si} isotope = x

  • The average atomic mass of silicon is 28.084 amu

By inserting these values into equation 1, we derive:

28.084=[(27.9769\times 0.9222)+(28.9764\times 0.0468)+(29.9737\times x)]\\\\x=0.0309

To convert this fractional abundance into a percentage, multiply by 100:

\Rightarrow 0.0309\times 100=3.09\%

This shows that the percentage abundance for _{14}^{30}\textrm{Si} isotope is 3.09 %.

6 0
3 months ago
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