Answer :
Some
of the solutions exhibit
colligative properties. These properties depend on the amount of solute
dissolved in a solvent. These properties include freezing point depression, boiling
point elevation, osmotic pressure and vapor pressure lowering. Calculations
are as follows:
ΔT(freezing point) = (Kf)mi
3 = 1.86 °C kg / mol (m)(2)
3 =3.72m
m = 0.81 mol/kg
Answer : The molality of a solution is, 0.806 mole/Kg
Explanation :
First we have to calculate the Van't Hoff factor (i) for KCl.
The dissociation of [tex]KCl[/tex] will be,
[tex]KCl\rightarrow K^++Cl^-[/tex]
So, Van't Hoff factor = Number of solute particles = [tex]K^++Cl^-[/tex] = 1 + 1 = 2
Now we have to calculate the molality of solution.
Formula used for lowering in freezing point :
[tex]\Delta T_f=i\times k_f\times m[/tex]
or,
[tex]T_f^o-T_f=i\times k_f\times m[/tex]
where,
[tex]\Delta T_f[/tex] = change in freezing point
[tex]T_f[/tex] = temperature of solution = [tex]-3^oC[/tex]
[tex]T^o_f[/tex] = temperature of pure water = [tex]0^oC[/tex]
[tex]k_f[/tex] = freezing point constant = [tex]1.86^oC/m[/tex]
m = molality = ?
i = Van't Hoff factor = 2
Now put all the given values in this formula, we get the molality of the solution.
[tex]0^oC-(-3^oC)=2\times (1.86^oC/m)\times m[/tex]
[tex]m=0.806mole/Kg[/tex]
Therefore, the molality of a solution is, 0.806 mole/Kg