Answer : The freezing point of the solution is, 260.503 K
Solution : Given,
Mass of methanol (solute) = 215 g
Mass of water (solvent) = 1000 g = 1 kg (1 kg = 1000 g)
Freezing depression constant = [tex]1.86^oC/m=1.86Kkg/mole[/tex]
Formula used :
[tex]\Delta T_f=K_f\times m\\T^o_f-T_f=K_f\times \frac{w_{solute}}{M_{solute}\times w_{solvent}}[/tex]
where,
[tex]T^o_f[/tex] = freezing point of water = [tex]100^oC=273K[/tex]
[tex]T_f[/tex] = freezing point of solution
[tex]K_f[/tex] = freezing point constant
[tex]w_{solute}[/tex] = mass of solute
[tex]w_{solvent}[/tex] = mass of solvent
[tex]M_{solute}[/tex] = molar mass of solute
Now put all the given values in the above formula, we get
[tex]273K-T_f=(1.86Kkg/mole)\times \frac{215g}{(32g/mole)\times (1kg)}[/tex]
By rearranging the terms, we get the freezing point of solution.
[tex]T_f=260.503K[/tex]
Therefore, the freezing point of the solution is, 260.503 K
From the calculations, the freezing point of the pure solution is - 12.5 °C.
The freezing point is the temperature at which the liquid water is turned to solid water.
In this case, we know that;
ΔT= [tex]K m i[/tex]
Since;
[tex]m = 215 g/32 g/mol/1 Kg = 6.7 m[/tex]
K = 1.86°C/mole
i = 1
Then
ΔT = [tex]1.86c/mole * 6.7 m * 1 = 12.5 C[/tex]
The freezing point of pure water is 0°C hence the freezing point of the solution is;
0°C - 12.5 °C = - 12.5 °C
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