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Salt Content in Potato Chips: Chemical Analysis

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Salt Content in Potato Chips: Chemical Analysis er en kemi-opgave til 1.g el. lign., afleveret til karakteren 4. Fylder 4 sider (1.286 ord, ca. 6 min. læsning) og blev publiceret 13. maj 2017.

This report details an experiment to determine the salt content in potato chips. It covers the theoretical background of dilution and titration, the experimental procedure, raw data processing, and an evaluation of the results. The experiment found a significantly higher salt content than the theoretical value, with a discussion of potential errors.

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Detailed lab report on determining salt content in chips. Includes theory, procedure, calculations, and a thorough error analysis, providing good inspiration.
Struktur
12
Faglig dybde
10
Kilder
7
Fuldstændighed
10
  • chemistry
  • dilution
  • error analysis
  • experimental report
  • lab report
  • potato chips
  • salt content
  • sodium chloride
  • titration

Chips are a very common snack all around the world. When we stand in the supermarket we can chose between a variety of flavors, but one thing that is in all of the chips is salt. The salt in the chips is creating an enjoyable taste, but too much salt, also known as sodium chloride, can cause increased blood pressure, which gives a higher risk of strokes and heart attacks. Therefore, it is important to know the salt content of chips. This salt content is usually labeled on the package of the chips, but you can also perform an experiment to find out yourself if not. To perform this experiment, you do the first step, which is a dilution of the chips in to water, and then stir it, and then the second step, which is a titration.

Theory

The potato chip experiment had the purpose of determining the amount of NaCl in 100g of chips. The first step to do this was to do a dilution. A dilution is adding more solvent, without adding more solute. To calculate the volume, that we need to add to a mixture to dilute we can use the formula: M1V1=M2V2. Here the M1 and V1, is the molar mass and volume, before adding anything, and M2 and V2 is the wished molar mass after dilution, and the volume after the dilution. For example (example taken from chemteam.info) we have 53.4 mL of a 1.50 M solution of NaCl, but we need a 0.800 M solution. How many mL of 0.800 M can you make? The calculations for this example will look like this: 1.5 mol L-1*53.4 mL=0.800 mol L-1*x

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