BIOS 105 lab 1

BIOS 105 lab 1

Separation of Saltwater Solution


The purpose of this lab is too separate the saltwater solution from an unknown mixture. In order for the saltwater to be separated from the mixture, the process called evaporation was being used. The water began to boil at a temperature of 90-100 degrees C, which then causes it evaporate in the air leaving only the Salt. From this, you can just easily weigh the remaining saltwater from the original weight of the unknown mixture and you can tell the weight of the water and the salt. In this experiment, I found that mass of salt by percentage is 10.84%.

Weigh the dry and empty 250 ml beaker in the electronic balance and record the data. Pour the unknown mixture in the 250 ml beaker and weigh again. Remove the beaker from the electronic balance and bring the unknown mixture to a boil until all the water had evaporated. Let it cool for 20 degrees C and weigh the beaker with the salt in it.

Observation and Result:

Mass of Beaker: 75.000 g
Mass of Beaker + Mixture: 85.175 g
Mass of Beaker + Salt: 75.995 g
Mass of Salt:
75.995 g – 75.000 g = 0.995 g
Mass of Water:
75.995 g – 85.175 g = 9.180 g
Temperature Water Began to boil : 90 C

Mass percent of salt:
(0.995 / 9.180) * 100 = 10.84%

In This experiment, I learned the easy way on how to calculate on how much salt was in saltwater mixture. Evaporation was the method being used in order for this experiment to be done. Water becomes gas when heated at 90 – 100 C and leaves whatever other mixture they have in it. I also found out that salt in a liquid form will turn to solid when heated at 100 C.

Water evaporates when heated at 100 C and whatever other mixture it has will be left behind. By this method we can tell how much water and salt mass was in a mixture. Also, accuracy in collecting data helps a lot in determining the results of each calculations, especially when all your record...

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