Respuesta :
(1.86 °c / 1 molal)×(2 molal) = 3.72°c, and because the temperature of the freezing point goes down the number has to be negative. So, the final answer is -3.72°c
Answer : The freezing point of the solution is, [tex]-3.72^oC[/tex]
Explanation : Given,
Molality of the solution = 2.00 m
Formula used :
[tex]\Delta T_f=i\times K_f\times m\\\\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^o[/tex] = temperature of pure water = [tex]0^oC[/tex]
[tex]T_f[/tex] = temperature of solution = ?
i = Van't Hoff factor for non-electrolyte solution = 1
[tex]K_f[/tex] = freezing point constant = [tex]1.86^oC/m[/tex]
m = molality = 2.00 m
Now put all the given values in this formula, we get:
[tex]0^oC-T_f=1\times 1.86^oC/m\times 2.00m[/tex]
[tex]T_f=-3.72^oC[/tex]
Therefore, the freezing point of the solution is, [tex]-3.72^oC[/tex]
