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1. Iodine is required for a healthy diet - Scottish Highers Chemistry - Question 11 - 2018

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1. Iodine is required for a healthy diet. Food grown in certain parts of the world is low in iodine. To prevent iodine deficiency in people’s diets, table salt can b... show full transcript

Worked Solution & Example Answer:1. Iodine is required for a healthy diet - Scottish Highers Chemistry - Question 11 - 2018

Step 1

Describe a procedure to accurately weigh out a 50.0 g sample of iodised table salt.

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Answer

To accurately weigh a 50.0 g sample of iodised table salt, use the following procedure:

  1. Place a clean, dry weighing boat or paper on a balance and tare it to zero.
  2. Carefully add iodised salt to the weighing boat until the display shows 50.0 g.
  3. Avoid spilling or losing any salt during the transfer, and ensure the balance is calibrated before weighing.

Step 2

The overall equation for the reaction of bromine solution with iodide ions is shown.

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Answer

The ion-electron equation for the oxidation reaction is:

ightarrow I₂(aq) + 2e⁻$$

Step 3

Calculate the average volume, in cm³, of sodium thiosulfate solution that should be used to determine the number of moles of iodine released.

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Answer

To calculate the average volume of sodium thiosulfate solution used, add the volumes from the three samples and divide by the number of samples:

Average Volume = ( \frac{10.0 + 9.4 + 9.6}{3} = \frac{29.0}{3} \approx 9.67 \text{ cm}^3 )

Step 4

Calculate the number of moles of iodine released from 50 cm³ of the standard salt solution.

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Answer

Using the average volume of sodium thiosulfate solution calculated previously (9.67 cm³), we can determine the moles of iodine released:

First, convert cm³ to dm³:

( 9.67 \text{ cm}^3 = 0.00967 \text{ dm}^3 )

Using the molarity of the sodium thiosulfate solution (0.0010 mol l⁻¹):

( ext{Moles of Na₂S₂O₃} = \text{Molarity} \times ext{Volume (dm³)} = 0.0010 \times 0.00967 = 9.67 \times 10^{-6} \text{ moles} )

From the stoichiometry of the reaction (1:2), the moles of iodine released are:

( 9.67 \times 10^{-6} \text{ moles of } I_2 = \frac{1}{2} \times 9.67 \times 10^{-6} = 4.85 \times 10^{-6} \text{ moles} )

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