Sir William Alexander Clarke Bustamante.
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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: Hợp chất CTHH 0 °C 10 °C 20 °C 30 °C 40 °C 50 °C 70 °C
Actini(III) hydroxide Ac(OH)3 0,0022
Amonia NH3 1176 900 702 565 428 333 188
Amoni azua NH4N3 16 25,3 37,1
View 42 more rows
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Answer : Option D) Hot and dry with low growing shrubs.
Explanation : According to the attached image of the Biome, this particular biome displays the hot and dry climatic region along with low growing shrubs. A biomes are the way to divide the largely occurring flora and fauna in a particular place on the Earth's surface. These divisions are made on the basis of climatic patterns, soil types, and the animals and plants that are found to inhabit in that area.