Answer:
The given parameter for the solubility of NaNO₃ in H₂O are;
The maximum solubility of NaNO₃ in H₂O at 10°C = 78 g
1. If we have 50 g of NaNO₃ in 100 g of water at 10°C, the solution is;
Undersaturated
2. If we have exactly 78 g of NaNO₃ in 100 g of water at 10°C the solution is;
A saturated solution
3. If will add 80 g of NaNO₃ in 100 g of water at 10°C,
The excess NaNO₃ which cannot be dissolved will be observed as solids in the mixture
4) The quantity of NaNO₃ that can be dissolved in H₂O at 40°C = 94 g
If the solution containing 94 g of NaNO₃ at 40°C is cooled to 10C about 16 g of NaNO₃ will precipitate out of the solution and exist as solids in the mixture
Explanation:
1. An undersaturated is a solution that holds smaller amount of solute that it can hold at a given temperature
2. A saturated solution holds as much solute as it can dissolve at a given temperature
3. When more solutes are added to a saturated solution, the excess solution will remain in the solid form in the mixture
4) Cooling a saturated solution below the saturation temperature will result in the precipitation or crystallization of the excess solutes in the solution.
There are lots of variables.
1. relative humidity
2. dew point temperature
3. air temperature
4. wind speeds
5. vapor pressure
6. latitude and longitude
7. orrographic lift (surface type)
there are probably more, but hope this helped! :)
Normal boiling point is 99.97 degrees C and 211.9 degrees ferinhight. And a pressure of 1atm or 101.325 kPa. Hope this helps :)
Here this is what happens
