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
Answer:
Vf = 1.22 mL
Explanation:
If we assume that the pressure is constant and the number of moles does not change, we can say that the volume of the gas is modified in a directly ratio, to the Absolute Temperature.
Let's convert the values:
91°C + 273 = 364K
0.9°C + 273 = 273.9K
Volume decreases if the temperature is decreases
Volume increases if the T° increases
V₁ / T₁ = V₂ / T₂ → 1.63mL /364K = V₂ / 273.9K
V₂ = (1.63mL /364K) . 273.9K → 1.22 mL
The answer would be D. because when an acid and base mix the start to cancel each other out causing it to neutralize
what??? i need more information
