Question

What mass of glycerin (C₃H₈O₃), a nonelectrolyte, must be dissolved in 200.0 g of water to give a solution with a freezing point of $\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{°}\mathbf{C}$?

Short answer

14.83g of glycerin is required to dissolve in 200.0g of water so that the freezing point of the solution will be $-1.50°\text{C}$.

Step-by-step solution

Step 1:

According to the question, it is given that

mass of solvent is 200.0g

Depression at freezing point = $-1.50°\text{C}$

${\text{K}}_{\text{f}}\text{=1.86}°\text{C/m}$

Molar mass of glycerin = $12\times \text{3+8}\times \text{1+16}\times \text{3=92g}$

i = 1 for non-electrolyte solute

To determine the mass of glycerin, we have to calculate the molality of the solution which can be determined by depression in the freezing point i.e.

$\begin{array}{l}0\text{-}\Delta {\text{T}}_{\text{f}}{\text{=k}}_{\text{f}}\times \text{m}\times \text{i}\\ 0\text{-(-1.50)=1.86}\times \text{m}\times \text{1}\\ \text{m=}\frac{1.50}{\text{1.86}}\text{=0.806m}\end{array}$

Step 2:

Now, by using the molality of a solution, we can calculate the mass of solute which is glycerine.

$\begin{array}{l}\text{Molality=}\frac{\text{Massofsolute}\times \text{1000}}{\text{Molarmassofsolute}\times \text{Massofsolvent}}\\ \text{0.806=}\frac{\text{Massofsolute}\times \text{1000}}{\text{92}\times \text{200}}\\ \text{Massofsolute=}\frac{\text{0.806}\times \text{92}\times \text{200}}{1000}\text{=14.83g}\end{array}$