Data in tables can be extremely useful and informative. Sometimes, though, making a graph of data makes relationships between variables much more clear than a table can. In this activity you will graph four sets of data on the same sheet of graph paper. These data describe the maximum solubility of four different substances in water as a function of temperature. Two of the substances are gases and two of the substances are solids.
Maximum solubility is also known as saturation. When a solution is saturated it means that the solvent has dissolved as much of the solute as it can and no more solute will dissolve. If more solute is added to a saturated solution it will not dissolve. If the solute is a solid it will remain visible in the solid phase. On the level of ordinary observation nothing appears to be happening. On the molecular level, particles of solid are constantly dissolving and exiting the solution to become solid again.
Here are the data you will graph:
Temperature (°C)  Grams of NaCl
Table Salt solid 
Grams of O_{2}
Oxygen gas 
Grams of NH_{3}
Ammonia gas 
Grams of C_{12}H_{22}O_{11}
Table Sugar (sucrose) solid 

0  34.5  14.0  92.2  64.4  
10  35.0  11.0  72.1  65.4  
20  35.4  8.8  57.0  66.7  
30  35.8  7.5  45.9  68.3  
40  36.2  6.5  37.8  70.1  
50  36.6  5.7  29.6  72.1  
60  37.0  5.0  24.5  74.2  
70  37.4  4.6  19.4  76.5  
80  37.8  4.0  15.3  78.7  
90  38.2  3.5  10.2  80.7  
100  38.6  3.1  8.1  82.6 
The table shows the number of grams of solute that will combine with 100 g of water to make a saturated solution of that solute. For example, 34.5 g of NaCl plus 100.0 g of water make up 134.5 g of solution and this would be the maximum amount that would dissolve at 0°C.
The numbers in the table can be manipulated to find out how much solute you would add to any amount of water to make a saturated solution. Volumes of water in mL can stand for mass of water since for low temperatures because the density of water is almost exactly 1 g/mL. Say you want to make a saturated solution with 50 g of water at 20°C:
Set up a proportion like this: 35.4 g salt x g salt  =  100 g water 50 g water 
this simplifies to: 50  × 35.4 g salt = 17.7 g 100 
So, to prepare the solution described you would mix 17.7 g of salt with 50.0 g of water. The total mass of the solution you make would be 67.7 g. The table relates how much of each solute will dissolve in 100 g water at each temperature. For other amounts of water you just set up a simple proportion.
Say you want to make 100 g of a saturated solution. From the table you can read that 100 g of water requires 35.4 g of salt to be saturated at 20°C. This makes a total solution mass of 135.4 g. That means that 135.4 g of solution contain 100 g water and 35.4 g salt. This is more than you want to make. In order to make a desired total mass for the solution you need to find the percent solute by mass. To calculate percentage by mass just divide and multiply by 100%:
35.4 g salt  × 100% = 26.14% 135.4 g solution
So the salt solution is 26.14% salt by mass. The other 73.86% of the solution’s mass is water. Now it’s easy to make 100 g of solution. Just measure 26.14 g of salt and add water until the total mass is 100 g. To make 50 g total mass for a saturated solution at 20°C just measure 0.2614(50) = 13.07 g of salt and add water until the total mass is 50 g.
To make your graph:
Answer these questions on a separate sheet of paper using complete sentences. Show work for all mathematical questions.
For this assignment you must turn in the answers to these questions and your sheet of graph paper showing all four plots. This will count toward your lab grade.