An aqueous solution is 10% glucose by mass $(d=1.039g/mLat{20}^{o}C)$. Calculate its freezing point, boiling point at 1atm, and osmotic pressure.

Molality is the amount of a substance dissolved in certain mass of solvent. The moles of a solute per kilogramme of a solvent are how it is defined.

Here, an aqueous solution of 10% glucose means 10 g of glucose dissolved in 100 mL (0.1kg) of water.

molar mass of glucose(C₆H₁₂O₆)

$M=6\times 12+12\times 1+6\times 16$=180g/mol.

Mass of solvent$=density\times volume=1.039g/mL\times 100mL=103.9g$

Hence, the moles of solute $=\frac{(mass\hspace{0.33em}of\hspace{0.33em}solute)}{(molar\hspace{0.33em}mass\hspace{0.33em}of\hspace{0.33em}solute)}$

$=\frac{10g}{(180g/mol)}=0.055moles$

Now, molality of solution$$\begin{gathered} =\frac{(moles\hspace{0.33em}of\hspace{0.33em}solute)}{(mass\hspace{0.33em}of\hspace{0.33em}solvent(kg)} \\ =0.055mol/0.01039kg=4.20mol/kg \end{gathered}$$

Freezing point of a substance is the temperature at which the vapor pressure of a substance in its liquid phase is equal the vapor pressure in solid phase.

To calculate it, use the formula below.

${\mathit{T}}_{\mathbf{f}}\mathbf{=}{\mathit{K}}_{\mathbf{f}}\mathbf{\times }\mathit{m}$

Where T_f is freezing point depression, $K_f(-1.{86}^{\circ }C/m)$is freezing point depression constant and m is molality.

Hence,

localid="1663341055294" $T_f=-1.{86}^{\circ }C/m\times 4.20mol/kg\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=-7.8^{\circ }C$

Therefore, the freezing point of solution is $7.8^{\circ }C$.

Boiling point is the temperature at which the pressure exerted by surroundings upon a liquid is equaled by the pressure exerted by the vapor of the liquid.

To calculate it, use the formula below.

${\mathit{T}}_{\mathbf{b}}\mathbf{=}{\mathit{K}}_{\mathbf{b}}\mathbf{\times }\mathit{m}$

Where T_bis the boiling point elevation, K_b(0.512C/m) is boiling point elevation constant and m is molality.

$T_b=0.{512}^{\circ }C/m\times 4.20mol/kg\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=2.1^{\circ }C$

Therefore, the solution's boiling point is 2.1^.C.

Osmotic pressure is defined as the minimum pressure that must be applied to halt the flow of solvent molecules through a semipermeable membrane. Utilizing the formula n=MRT, it may be calculated. Where M is the molarity of solution, R gas constant and T temperature.

Given temperature$={20}^{\circ }C=20+273=293K$

Value of R is$0.082L.atm/mol.K$

Hence, osmotic pressure

$\Pi =4.20mol/kg\times 0.082L.atm/mol.K\times 293K\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=100.90atm$