Answer:
The freezing point will be 
Explanation:
The depression in freezing point is a colligative property.
It is related to molality as:

Where
Kf= 
the molality is calculated as:




Depression in freezing point = 
The new freezing point = 
Answer:
The plane with aluminium can lift more mass of passangers than the plane of steel.
Explanation:
The total mass the airplane canc lift is:

For aluminium:


and
![V_{fuselage}=\frac{\pi *L}{4}*[D^2-(D-e)^2]](https://tex.z-dn.net/?f=V_%7Bfuselage%7D%3D%5Cfrac%7B%5Cpi%20%2AL%7D%7B4%7D%2A%5BD%5E2-%28D-e%29%5E2%5D)
where:
- L is lenght
- D is diameter
- e is thickness
![m_{tot}=\delta _{Al}*\frac{\pi *L}{4}*[D^2-(D-e)^2]+m_{pas-Al}](https://tex.z-dn.net/?f=m_%7Btot%7D%3D%5Cdelta%20_%7BAl%7D%2A%5Cfrac%7B%5Cpi%20%2AL%7D%7B4%7D%2A%5BD%5E2-%28D-e%29%5E2%5D%2Bm_%7Bpas-Al%7D)
For steel (same procedure):
![m_{tot}=\delta _{Steel}*\frac{\pi *L}{4}*[D^2-(D-e)^2]+m_{pas-Steel](https://tex.z-dn.net/?f=m_%7Btot%7D%3D%5Cdelta%20_%7BSteel%7D%2A%5Cfrac%7B%5Cpi%20%2AL%7D%7B4%7D%2A%5BD%5E2-%28D-e%29%5E2%5D%2Bm_%7Bpas-Steel)
Knowing that the total mass the airplane can lift is constant and that aluminum has a lower density than the steel, we can afirm that the plane with aluminium can lift more mass of passangers.
Also you can estimate an average weight of passanger to estimate a number of passangers it can lift.
The control group in this experiment is the one with just distilled water. It is plain so they can use it to compare the other tests against.