Explanation:
A gas has a temperature of 273.15 K and a pressure of 101.325 kPa. It can be concluded that this gas has reached standard temperature and pressure.
Standard temperature is zero degree celcius which corresponds to 273.15 degree kelvin.
Standard pressure is 760 mmHg which corresponds to 101.325 kPa.
Answer : The molal freezing point depression constant of liquid X is, 
Explanation : Given,
Mass of urea (solute) = 5.90 g
Mass of liquid X (solvent) = 450 g = 0.450 kg
Molar mass of urea = 60 g/mole
Formula used :

where,
= change in freezing point
= freezing point of solution = 
= freezing point of liquid X = 
i = Van't Hoff factor = 1 (for non-electrolyte)
= Molal-freezing-point-depression constant = ?
m = molality
Now put all the given values in this formula, we get


Therefore, the molal freezing point depression constant of liquid X is, 
It is B: Fixed volume, takes the shape of the container
Explanation:
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