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Annette [7]
3 years ago
11

When standardizing an iodine solution, 30.37 ml of the solution was required to completely oxidize 10.00 ml of a 5 mg/ml solutio

n of ascorbic acid. what is the molarity of the titrant??
Chemistry
1 answer:
Svetach [21]3 years ago
4 0
First, we need to determine the reaction that happens when standardizing. The reactions would be:

 KIO3(aq) + 6 H+ (aq) + 5 I- (aq) ------> 3 I2(aq) + 3 H2O(l) + K+ (aq)
C6H8O6(aq) + I2(aq) ------> C6H6O6(aq) + 2 I- (aq) + 2 H+ (aq)

5g/L C6H8O6 (0.01 L) ( 1 mol / 176.12 g) ( 1 mol I2 / 1 mol C6H8O6 ) ( 1 mol KIO3 / 3 mol I2 ) = 0.00009 mol KIO3

Molarity = 0.00009 mol KIO3 / 0.03037 L = 0.003 M

Hope this answers the question. Have a nice day.
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Answer:

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Explanation:

Pipet is used to dispense a very small amount of liquid.

Test tube rack is used to hold multiple test tubes at the same time.

Test Table is used to view chemical reactions or hold or heat small amounts of substance.

Scoopula is used to dispense chemicals from a larger container.

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3 years ago
Best example of heterogeneous mixture
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A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s
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<u>Answer:</u> The percent yield of the 1-bromobutane is 48.65 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For NaBr:</u>

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol

The chemical equation for the reaction of 1-butanol and NaBr is:

\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = \frac{1}{1}\times 0.108=0.108 moles of 1-bromobutane

  • Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g

  • To calculate the percentage yield of 1-bromobutane, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%

Hence, the percent yield of the 1-bromobutane is 48.65 %

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