NY Chemistry Regents Exam June 2019
At 23°C, 85.0 grams of NaNO3(s) are dissolved in 100. grams of H2O(l).
55 Convert the temperature of the NaNO3(s) to kelvins. 
Solution: The formula for conversion is °C+273 = K
23°C+273 = 296 K
Answer: 296 K
56 Based on Table G, determine the additional mass of NaNO3(s) that must be dissolved to saturate the solution at 23°C. 
Solution: On Table G we must find the NaNO3 line and then find 23°C on the x axis. We then go up from the temperature until we reach the NaNO3 line. Looking on the y-axis we find the grams, which is around 90 grams. 90g-85g= 5g
Answer: Any number from 4g to 6g is accepted as the correct answer.
57 State what happens to the boiling point and freezing point of the solution when the solution is diluted with an additional 100. grams of H2O(l). 
Solution: When solute is added to pure water, the boiling point increases and freezing point decreases. Diluting the solution by adding more water results in a decreased concentration of solute, which would result in lower boiling point and higher freezing point.
Answer: lower boiling point and higher freezing point.