What is the freezing point of a solution of 210.0 g of glycerol, formula C3H8O3, dissolved in 350. g of water? Careful. First get molar mass and use molar mass to determine molality concentration. Then use freeze pt. depression formula to determine the change in freezing pt. Then determine the new freeze point. The freezing point depression constant for water is Kf= -1.86 oCelcius/molal. Report your answer rounded to 1 decimal point and do not include units.

The change in freezing point and the new freeze point are -12.1°C and -12.1°C.

Given the following data:

We know that the temperature at which water freezes is 0°C.

To determine the change in freezing point and the new freeze point:

First of all, we would determine the molar mass of glycerol:

Molar mass of glycerol () =

Molar mass of glycerol () =

Molar mass of glycerol () = 92 g/mol.

Next, we would find the number of moles of glycerol that is required:

Number of moles = 2.28 moles

To find the molality concentration:

Molality = 6.51 molal.

Mathematically, the freezing point elevation of a liquid is given by the formula:

Where:

Substituting the parameters into the formula, we have;

Change in temperature = -12.1°C.

Now, we can determine the new freeze point:

New freeze point = -12.1°C.

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- 12.1

Explanation:

M(C3H8O3) = 3*12.0 + 8*1.0 + 3*16.0 = 92 g/mol

210.0 g C3H8O3 * 1 mol C3H8O3/92 g C3H8O3 = 210/92 mol C3H8O3

350.g = 0.350 kg H2O

Molality = mol soluty/ kg solvent = (210/92 mol C3H8O3) /(0.350 kg H2O) = =6. 522 molal

ΔT =i* Kf* m

(T2 - T1) = i* Kf* m

(T2 - 0°C) = 1*(-1.86°C/molal *6.522 molal)

T2= - 12.1°C

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