Molar Solutions
-A molar solution is a solution in which one mole of a compound is dissolved in a total volume of 1 litre. One example is sodium chloride, which is 58.44 which is 1M NaCl solution. First weigh 58.44 grams of NaCl and dissolve it into one litre of water, or make a 1: 10 of a 1M sample.
-A molar solution is a solution in which one mole of a compound is dissolved in a total volume of 1 litre. One example is sodium chloride, which is 58.44 which is 1M NaCl solution. First weigh 58.44 grams of NaCl and dissolve it into one litre of water, or make a 1: 10 of a 1M sample.
![Picture](/uploads/2/3/4/2/23426136/9485303.jpg?351)
Materials:
-One beaker (200 mL)
-5 small cups
-171 grams of sugar
-1000 mL of water
-5 potatoes
- one scale
- 5 sheets of wax paper
-5 spoons
-sharpie
Protocol:
1. Gather all the materials
2. Labeled all the cups with the molar solution sucrose amount and team number (4)
3. Contained the DI cup and filled the beaker with 200 mL and poured the water into the DI cup
4. Gathered the 0.3 M cup and massed 20.52 sugar on the electronic scale and mixed the water and poured the sugared water into the cup.
5. Repeat step 4 for the cup of 0.5 M but with 34.2 of sucrose , 0.7 M with 47.88 amount of sucrose, and 1.0 M with 68.4 of sucrose . ( remember to be precise)
6.After all cups have their sugared water, 5 potatoes were contained and massed on the electronic scale
7. Once the 5 potatoes were massed and recorded, they were put into one of the 5 cups and recorded as well
8. After the potatoes were put into the cups the wax paper was contained and put on top of the cups and to hold them down spoons were put on top
9. Left the 5 cups with the potatoes in them for 24 hours
10. After the potatoes were left in the cups after one whole day we took out the potatoes from the specific cup and were cleaned making sure that the potatoes were dry
11. Once each potato was dry they were massed on the electronic scale and recorded
12.When the data was obtained we had to find the difference for each potato with the equation of Mf -MI and determined if their was a gain or a loss
13. Then we had to determine the % mass change with the equation of (MF-MI)/MI x 100
14. A graph had to be done with the molar solution sucrose and the % mass change
15. From the graph the molarity of the potato cell had been determined and in this case the molarity was .3 M
-One beaker (200 mL)
-5 small cups
-171 grams of sugar
-1000 mL of water
-5 potatoes
- one scale
- 5 sheets of wax paper
-5 spoons
-sharpie
Protocol:
1. Gather all the materials
2. Labeled all the cups with the molar solution sucrose amount and team number (4)
3. Contained the DI cup and filled the beaker with 200 mL and poured the water into the DI cup
4. Gathered the 0.3 M cup and massed 20.52 sugar on the electronic scale and mixed the water and poured the sugared water into the cup.
5. Repeat step 4 for the cup of 0.5 M but with 34.2 of sucrose , 0.7 M with 47.88 amount of sucrose, and 1.0 M with 68.4 of sucrose . ( remember to be precise)
6.After all cups have their sugared water, 5 potatoes were contained and massed on the electronic scale
7. Once the 5 potatoes were massed and recorded, they were put into one of the 5 cups and recorded as well
8. After the potatoes were put into the cups the wax paper was contained and put on top of the cups and to hold them down spoons were put on top
9. Left the 5 cups with the potatoes in them for 24 hours
10. After the potatoes were left in the cups after one whole day we took out the potatoes from the specific cup and were cleaned making sure that the potatoes were dry
11. Once each potato was dry they were massed on the electronic scale and recorded
12.When the data was obtained we had to find the difference for each potato with the equation of Mf -MI and determined if their was a gain or a loss
13. Then we had to determine the % mass change with the equation of (MF-MI)/MI x 100
14. A graph had to be done with the molar solution sucrose and the % mass change
15. From the graph the molarity of the potato cell had been determined and in this case the molarity was .3 M
Analysis
![Picture](/uploads/2/3/4/2/23426136/1380665346.png)
Molarity is basically the amount of solute dissolved in the solvent. The graph shows that the molarity of the potato was .3 because fit is the one that crossed the x-axis. Once it was 0.5 Molar their was loss. From the set up at the beginning it explains what molarity is because it consist of massing, and the solute in a solvent. The lab set up also defines what molarity is because in the lab we had a lot of massing sucrose which was mixed in with water and made a solution. After that we put in potatoes which were in the cups and through time with massing we figured out how much sucrose the potato had contained and the molarity of it. The experiment show that the potato can contain but it can lose as well after it has reached its molarity.