Question

3.Calculate the freezing point of a solution containing 60g of glucose (Molar mass=180
g/mol) in 250g of water.(Kfof water = 1.86 Kkg mol-1)​

Answer 1

Answer:

$\bigstar{\bold{Freezing\:point\:of\:solution=270.67\:K}}$

Explanation:

$\Large{\underline{\underline{\rm{Given:}}}}$

• Mass of solute-glucose (w₂) = 60 g
• Molar mass of solute (M₂) = 180 g/mol
• Mass of solvent (w₁) = 250 g
• Molar mass of solvent = 2 + 16 = 18 g/mol
• $\sf{K_f}$ of water = 1.86 K kg/mol

$\Large{\underline{\underline{\rm{To\:Find:}}}}$

• Freezing point of the solution

$\Large{\underline{\underline{\rm{Solution:}}}}$

↬ Here we have to find the freezing point of the solution

↬ First we have to find the freezing point depression of the solution

↬ The freezing point depression is given by,

    $\sf{M_2=\dfrac{K_f\times w_2\times 1000}{\Delta T_f\times w_1}}$

↬  Substitute the data, we get the value of $\sf{\Delta T_f}$,

     $\sf{180\:g/mol=\dfrac{1.86\:K\:kg/mol\times 60\:g\times 1000}{\Delta\:T_f\times 250g} }$

↬ Now solving it we get the value of depression of freezing point,

   $\sf{180=\dfrac{111600}{\Delta T_f\times 250} }$

   $\sf{\Delta T_f=\dfrac{111600}{45000} }$

   $\sf{\Delta T_f=2.48\:K}$

↬ Hence the depression in freezing point is 2.48 K

↬ Now the normal freezing point of water is 273.15 K

↬ Hence the freezing point of the solution is given by,

  Freezing point = Freezing point of solvent - Depression in freezing   point

↬ Substitute the data,

   Freezing point = 273.15 K - 2.48 K

   Freezing point = 270.67 K

↬ Hence the freezing point of the solution is 270.67 K

   $\boxed{\bold{Freezing\:point\:of\:solution=270.67\:K}}$

$\Large{\underline{\underline{\rm{Notes:}}}}$

↬ The freezing point depression constant is called as the Molal depression constant or Cryoscopic constant.

↬ The unit of $\sf{K_f}$ is K kg/mol

↬ Depression in freezing point is related to the molality of the solution.

    $\sf{\Delta T_f\propto \:m}$

    $\sf{\Delta T_f=K_f\:m}$