Respuesta :
Explanation:
When a non-volatile solute is added in a solvent then decrease in its freezing point is known as freezing point depression.
Mathematically, [tex]\Delta T = K_{f}m[/tex]
where [tex]\Delta T[/tex] = change in freezing point
[tex]K_{f}[/tex] = freezing point depression constant [tex]\text({in ^{o}C/mol/kg})[/tex]
m = molality
First, calculate the number of moles as follows.
No. of moles = [tex]\frac{\text{mass of solute}}{\text{molar mass of solute}}[/tex]
= [tex]\frac{22.78g}{62.07 g/mol}[/tex]
= 0.367 mol
Now, it is given that mass of solvent is 87.95 g. As there are 1000 grams in 1 kg.
So, [tex]87.95g \times \frac{1 kg}{1000 g}[/tex] = 0.08795 kg
Hence, molality of the given solution is as follows.
Molality = [tex]\frac{\text{no. of moles}}{\text{mass in kg}}[/tex]
= [tex]\frac{0.367 mol}{0.08795}[/tex]
= 4.172 mol/kg
Therefore, depression in freezing point will be as follows.
[tex]\Delta T = K_{f}m[/tex]
[tex]T_{solvent} - T_{mixture}[/tex] = [tex]K_{f}m[/tex]
Since, freezing point of pure water is [tex]0^{o}C[/tex]. Now, putting the given values as follows.
[tex]0^{o}C - T_{mixture}[/tex] = [tex]1.86^{o}C/m \times 4.172mol/kg[/tex]
= [tex]-7.759 ^{o}C[/tex]
Thus, we can conclude that the freezing point of water in the given mixture is [tex]-7.759 ^{o}C[/tex].
