Answer :
Answer : The freezing depression of a solution is [tex]-1.65^oC[/tex]
Explanation : Given,
Molal-freezing-point-depression constant [tex](K_f)[/tex] for water = [tex]1.86^oC/m[/tex]
Mass of glycerin (solute) = 30.7 g
Volume of water (solvent) = 376 mL
Molar mass of glycerin = 92.09 g/mole
First we have to calculate the mass of water.
As we know that, the density of water is 1.0 g/mL.
So, mass of water = Density of water × Volume of water
Mass of water = 1.0 g/mL × 376 mL
Mass of water = 376 g = 0.376 kg (1 g = 0.001 kg)
Now we have to calculate the freezing point depression of a solution.
Formula used :
[tex]\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of glycerin}}{\text{Molar mass of glycerin}\times \text{Mass of water in Kg}}[/tex]
where,
[tex]\Delta T_f[/tex] = change in freezing point
[tex]\Delta T_s[/tex] = freezing point of solution = ?
[tex]\Delta T^o[/tex] = freezing point of water = [tex]0^oC[/tex]
i = Van't Hoff factor = 1 (for glycerin non-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]0^oC-T_s=1\times (1.86^oC/m)\times \frac{30.7g}{92.09g/mol\times 0.376kg}[/tex]
[tex]T_s=-1.65^oC[/tex]
Therefore, the freezing depression of a solution is [tex]-1.65^oC[/tex]
Answer:
The freezing point depression of a solution is [tex]\text\bold -1.65^{o} C[/tex].
Explanation:
It is given that,
Given mass of glycerin is = 30.7 grams (Solute)
Volume of water = 376 mL
[tex]\text{K}_\text{f}[/tex] or molal-freezing-depression point is = [tex]1.86^oC/m[/tex]
Molar mass of glycerin = 92.09 g/mole
Now, to determine the [tex]\text{K}_\text{f}[/tex] value, the mass of water should be known. Thus, to calculate, the formula used will be:
Mass = Density X Volume
Mass = 1.0 g/mL X 376 mL
Mass = 376 g or 0.376 Kg
Using the formula of freezing point depression, the equation becomes:
[tex]\Delta\text {T}_\text{f} &=& \text i\;*\;\text K_{f} \;*\;\text m\\\text T^{o} - \text T_{s} &=& \text i\;*\;\text K_{f}\;*\; \frac{\text{Mass of glycerin}}{\text{Molar mass of glycerin * Mass of water in Kg}}[/tex]
in which,
[tex]\Delta\text {T}_\text{f}[/tex] = change in freezing point
[tex]\Delta\text {T}_\text{s}[/tex] = freezing point of solution that has to be find
[tex]\Delta\text {T}^{o}[/tex] = freezing point of water ([tex]0^{o} C[/tex])
Since, glycerin is a non-electrolyte, the Van't Hoff factor will be 1.
Substituting the values in the above equation:
[tex]\text {0}^\text{o}C - T_{s} &=& \text 1\;*\;\text (1.86^{o} C/m) \;*\;\frac{{30.7}}{92.09g/mol*0.376\text Kg} \\\text T_{s} = -1.65^{o} C[/tex]
Thus, the freezing point depression of a solution is [tex]-1.65^{o} C[/tex].
For Further Reference:
https://brainly.com/question/16694533?referrer=searchResults