Question

The data below was collected in an experiment to determine the solubility of sodium nitrate at 20 Celsius degree.

Total Volume of Water (mL)	Saturation Temp (Celsius degree)	Solubility (g salt/100g water)
3.50	78.0
3.75	70.0
4.00	60.0
4.25	52.0
4.50	45.0
4.75	40.0
5.00	36.0

1. If 5.0000g of NaNO3 was used, calculate the solubility in units of g NaNO3/100g water at each saturation temperature. Show your first calculation. Complete the rest of the calculations and fill in the table.

2. Construct a graph of solubility as a function of saturation.

3. Determine the solubility of sodium nitrate at 20 Celsius degree from a graph.

4. Using the solubility from #3, calculate the percent by mass of the salt in a saturated solution at 20 Celsius degree.

5. If the density of a saturated solution of sodium nitrate at 20 Celsius degree is found to be 1.4g/mL, calculate the Molarity of the solution.

Answer 1

We can calculate the solubility if we find out the amount of salt that will dissolve in 100 g of water, knowing that 5g dissolve in a given volume, (the density of water at 20ºC is 0,998 g/mL), so at 78ºC:

we repeat this procedure for each temperauture and graph the data,

we add a tendency line to extend the behaviour and estimate the solubility of the salt at 20ºC,

finally, we can calculate the molarity of a saturated solution at 20ºC

remember that the mass of the solution will be the sum of the mass of solute (85.66g) plus the mass of solvent (100 g), this way the mass of the solution is 185.66 g