Answer : The molal freezing point depression constant of X is 
Explanation : Given,
Mass of urea (solute) = 5.90 g
Mass of X liquid (solvent) = 450.0 g
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 of X = ?
m = molality
Now put all the given values in this formula, we get
![[0.4-(-0.5)]^oC=1\times k_f\times \frac{5.90g\times 1000}{60g/mol\times 450.0g}](https://tex.z-dn.net/?f=%5B0.4-%28-0.5%29%5D%5EoC%3D1%5Ctimes%20k_f%5Ctimes%20%5Cfrac%7B5.90g%5Ctimes%201000%7D%7B60g%2Fmol%5Ctimes%20450.0g%7D)
Therefore, the molal freezing point depression constant of X is
Water can change it's STATE but it will not change it's COMPOUND.
Answer:
982.5 kg/m³
Explanation:
When the temperature of a fluid increases, it dilates, and because of the variation of the volume, it's density will vary too. The density can be calculated by the expression:
ρ₁ = ρ₀/(1 + β*(t₁ - t₀))
Where ρ₁ is the final density, ρ₀ the initial density, β is the constant coefficient of volume expansion, t₁ the final temperature, and t₀ the initial temperature.
At t₀ = 4°C, the water desity is ρ₀ = 1,000 kg/m³. The value of the constant for water is β = 0.0002 m³/m³ °C, so, for t₁ = 93°C
ρ₁ = 1,000/(1 + 0.0002*(93 - 4))
ρ₁ = 1,000/(1+ 0.0178)
ρ₁ = 982.5 kg/m³
So what am I suppose to answer here?
