Respuesta :
Answer:
Depression in freezing point = 2 X 1.853 X 0.25 = 0.9625
Thus this will be the difference between the freezing point of pure water and the solution.
Explanation:
On adding any non volatile solute to a solvent its boiling point increases and its freezing point decreases [these are two of the four colligative properties].
The depression in freezing point is related to molality of solution as:
ΔTf[tex]=iK_{f}Xmolality[/tex]
where
ΔTf= depression in freezing point
Kf= cryoscopic constant of water = 1.853 K. kg/mol.
i = Van't Hoff factor = 2 ( for KCl)
molality = [tex]\frac{molesofsolute}{massofsolvent(Kg)}[/tex]
moles of solute = mass / molarmass = 4.66 / 74.55 =0.0625
mass of solvent = mass of solution (almost)
considering the density of solution to be 1g/mL
mass of solvent = 250 grams = 0.250 Kg
molality = [tex]\frac{0.0625}{0.25}= 0.25[/tex]
Putting values
depression in freezing point = 2 X 1.853 X 0.25 = 0.9625
Thus this will be the difference between the freezing point of pure water and the solution.
Answer : The solution will be [tex]0.931^oC[/tex] lower than water.
Explanation : Given,
Molal-freezing-point-depression constant [tex](K_f)[/tex] for water = [tex]1.86^oC/m[/tex]
Mass of KCl (solute) = 4.66 g
Volume of water = 250 mL
Density of water = 1.00 g/mL
So,
Mass of water (solvent) = [tex]Density\times volume=1.00g/mL\times 250mL=250g=0.250kg[/tex]
Molar mass of KCl = 74.5 g/mole
Formula used :
[tex]\Delta T_f=i\times K_f\times m\\\\\Delta T=i\times K_f\times\frac{\text{Mass of KCl}}{\text{Molar mass of KCl}\times \text{Mass of water in Kg}}[/tex]
where,
[tex]\Delta T[/tex] = change in freezing point
i = Van't Hoff factor = 2 (for KCl electrolyte)
[tex]K_f[/tex] = freezing point constant for water = [tex]1.86^oC/m[/tex]
m = molality
Now put all the given values in this formula, we get
[tex]\Delta T=2\times (1.86^oC/m)\times \frac{4.66g}{74.5g/mol\times 0.250kg}[/tex]
[tex]\Delta T=0.931^oC[/tex]
[tex]T^o-T_s=0.931^oC[/tex]
From this we conclude that, the solution will be [tex]0.931^oC[/tex] lower than water.
Therefore, the solution will be [tex]0.931^oC[/tex] lower than water.
