Respuesta :
Answer: The freezing point of solution is -0.974°C
Explanation:
- To calculate the concentration of solute, we use the equation for osmotic pressure, which is:
[tex]\pi=iMRT[/tex]
where,
[tex]\pi[/tex] = osmotic pressure of the solution = 12.1 atm
i = Van't hoff factor = 1 (for non-electrolytes)
M = molarity of solute = ?
R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
T = temperature of the solution = 298 K
Putting values in above equation, we get:
[tex]12.1atm=1\times M\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\M=\frac{12.1}{1\times 0.0821\times 298}=0.495M[/tex]
This means that 0.495 moles of glucose is present in 1 L or 1000 mL of solution
- To calculate the mass of solution, we use the equation:
[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]
Density of solution = 1.034 g/mL
Volume of solution = 1000 mL
Putting values in above equation, we get:
[tex]1.034g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.034g/mL\times 1000mL)=1034g[/tex]
- To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of glucose = 0.495 moles
Molar mass of glucose = 180.16 g/mol
Putting values in above equation, we get:
[tex]0.495mol=\frac{\text{Mass of glucose}}{180.16g/mol}\\\\\text{Mass of glucose}=(0.495mol\times 180.16g/mol)=89.18g[/tex]
Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.
- The equation used to calculate depression in freezing point follows:
[tex]\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}[/tex]
To calculate the depression in freezing point, we use the equation:
[tex]\Delta T_f=iK_fm[/tex]
Or,
[tex]\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}[/tex]
where,
Freezing point of pure solution = 0°C
i = Vant hoff factor = 1 (For non-electrolytes)
[tex]K_f[/tex] = molal freezing point elevation constant = 1.86°C/m
[tex]m_{solute}[/tex] = Given mass of solute (glucose) = 89.18 g
[tex]M_{solute}[/tex] = Molar mass of solute (glucose) = 180.16 g/mol
[tex]W_{solvent}[/tex] = Mass of solvent (water) = [1034 - 89.18] g = 944.82 g
Putting values in above equation, we get:
[tex]0-\text{Freezing point of solution}=1\times 1.86^oC/m\times \frac{89.18\times 1000}{180.16g/mol\times 944.82}\\\\\text{Freezing point of solution}=-0.974^oC[/tex]
Hence, the freezing point of solution is -0.974°C
