At 298 K, the osmotic pressure of a glucose solution (C6H12O6 (aq)) is 12.1 atm. Calculate the freezing point of the solution. The density of the solution is 1.034 g/mL.

The freezing point of solution is -0.974°C

Explanation:

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

Density of solution = 1.034 g/mL

Volume of solution = 1000 mL

Putting values in above equation, we get:

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.

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

the answer is option a.