Answer:
Percent composition of the solution is 26 % of sucrose and 74 % of water
Explanation:
Percent composition is the mass of solute, either of solvent in 100 g of solution.
Mass of solution = Mass of solvent + Mass of solute
Mass of solute = 35 g
Mass of solvent = 100 g
As we know, water density = 1g/mL
So 1g/mL . 100 mL = 100 g
35 g + 100 g = 135 g → Mass of solution
(Mass of solute / Mass of solution) . 100 =
(35 g / 135 g) . 100 = 26 %
(Mass of solvent / Mass of solution) . 100 =
(100 g / 135 g) . 100 = 74 %
Answer:
a
Explanation:
it is an electrolyte because of its strong polar chemical bond
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
Molality=mol/kg
342/171
=2m
Hope this helped :)
