The normal boiling point corresponds to the temperature at which both the liquid and gas are under the standard atmospheric pres
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<u>Answer:</u> The freezing point of solution is -0.974°C
<u>Explanation:</u>
- To calculate the concentration of solute, we use the equation for osmotic pressure, which is:
where,
= osmotic pressure of the solution = 12.1 atm
i = Van't hoff factor = 1 (for non-electrolytes)
M = molarity of solute = ?
R = Gas constant =
T = temperature of the solution = 298 K
Putting values in above equation, we get:
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:
Density of solution = 1.034 g/mL
Volume of solution = 1000 mL
Putting values in above equation, we get:
- To calculate the number of moles, we use the equation:
Moles of glucose = 0.495 moles
Molar mass of glucose = 180.16 g/mol
Putting values in above equation, we get:
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:
To calculate the depression in freezing point, we use the equation:
Or,
where,
Freezing point of pure solution = 0°C
i = Vant hoff factor = 1 (For non-electrolytes)
= molal freezing point elevation constant = 1.86°C/m
= Given mass of solute (glucose) = 89.18 g
= Molar mass of solute (glucose) = 180.16 g/mol
= Mass of solvent (water) = [1034 - 89.18] g = 944.82 g
Putting values in above equation, we get:
Hence, the freezing point of solution is -0.974°C