Question

The freezing point of aqueous solution that contains $5\%$ mass urea, $1.0\%$ mass $\mathrm{KCl}$ and $10\%$ mass of glucose is-

Answer 1

Freezing point depression

• A solution's freezing point is lower than that of the pure solvent.
• This indicates that freezing can only take place when a solution is lowered to a lower temperature than the pure solvent.

Step 1: Analyzing the given values

• The aqueous solution comprises $10\%$ by mass of glucose, $1.0\%$ by mass of $\mathrm{KCl}$, and $5\%$ by mass of urea.
• Assume that the solution weighs $100\mathrm{g}$. It will have $10\mathrm{g}$ of glucose, $1\mathrm{g}$ of $\mathrm{KCl}$, and $5\mathrm{g}$ of urea.
• Urea, $\mathrm{KCl}$, and glucose have molar weights of $60\mathrm{g}{\mathrm{mol}}^{-1},74.5\mathrm{g}{\mathrm{mol}}^{-1},\mathrm{and}180\mathrm{g}{\mathrm{mol}}^{-1}$, respectively.

Step 2: Calculating moles

• The number of moles of urea =$\frac{5\mathrm{g}}{60\mathrm{g}{\mathrm{mol}}^{-1}}$
• The number of moles of $\mathrm{KCl}$ = $\frac{1\mathrm{g}}{74.5\mathrm{g}{\mathrm{mol}}^{-1}}$
• The number of moles of glucose = $\frac{10\mathrm{g}}{180\mathrm{g}{\mathrm{mol}}^{-1}}$

Step 3: Calculating molality

• The quantity of solute particles affects the freezing point depression. Urea and glucose aren't electrolytes. Consequently, they don't decay.
• A powerful electrolyte is $\mathrm{KCl}$ . In an aqueous solution, it breaks down into K+ ions and Cl- ions. Therefore, multiply the amount of moles of $\mathrm{KCl}$ by $2$.
• Solvent mass is $100-[5+1+10]=84\mathrm{g}$.
• The ratio of the total moles of solute particles to the mass of solvent in kilogrammes is known as the molality of the solution.
• $\mathrm{Molality}\mathrm{m}=\frac{{\displaystyle \frac{5\mathrm{g}}{60\mathrm{g}{\mathrm{mol}}^{-1}}}+2\times {\displaystyle \frac{1\mathrm{g}}{74.5\mathrm{g}{\mathrm{mol}}^{-1}}}+{\displaystyle \frac{10\mathrm{g}}{180\mathrm{g}{\mathrm{mol}}^{-1}}}}{84\mathrm{g}}\times 1000=1.97\mathrm{m}$

Step 4: calculating depression in freezing point

• Decline in the freezing point ${\mathrm{\Delta T}}_{\mathrm{f}}={\mathrm{K}}_{\mathrm{f}}\mathrm{m}=1.86\times 1.97=3.66$
• Water has a freezing point of $273.15\mathrm{K}$
• Aqueous solution have a freezing point of $273.15-3.66=269.48\mathrm{K}.$

Thus, freezing point of the solution is $269.48\mathrm{K}$