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
Kinetic energy=Ek
Ek=(1/2)mv²
Ek=480 J
v=8 m/s
mass=?
Ek=(1/2)mv²
480 J=(1/2)m(8 m/s)²
480 J=(32 m²/s²) m
m=(480 J)/(32 m²/s²)=15 kg
answer: the mass of the object is 15 kilograms.
Answer:
1.3 M.
Explanation:
- We need to calculate the mass of the solution:
mass of the solution = mass of MgCl₂ + mass of water
mass of MgCl₂ = 20.1 g.
mass of water = d.V = (157.0 mL)(1.0 g/cm³) = 157.0 g.
∴ mass of the solution = mass of MgCl₂ + mass of water = 20.1 g + 157.0 g = 177.1 g.
- Now, we can get the volume of the solution:
V of the solution = (mass of the solution)/(density of the solution) = (177.1 g)/(1.089 g/cm³) = 162.62 mL = 0.163 L.
Molarity is the no. of moles of solute dissolved in a 1.0 L of the solution.
M = (no. of moles of MgCl₂) / (Volume of the solution (L)).
<em>∴ M = (mass/molar mass)of MgCl₂ / (Volume of the solution (L)) =</em> (20.1 g/95.211 g/mol) / (0.163 L) = <em>1.29 M ≅ 1.3 M.</em>
