Question

What would the final freezing point of water be if 3 mol of sugar were added to 1 kg of water (Kf = 1.86C/(mol/kg) for water and i = 1 for sugar)?

Answer 1

Answer:

The final freezing point of water is -5.58$^{0}C$

Explanation:

According to colligative properties of molecules, $\Delta T_{f}=i.k_{f}.m$

where $\Delta T_{f}$ is the depression in freezing point of a solution, i is the vant hoff factor of solute, $k_{f}$ is the cryogenoscopic constant of solvent and m is molality of the solution.

Here solute is sugar and solvent is water.

Molality of solution = $\frac{number of moles of sugar}{mass of water in kg}$ = $\frac{3}{1}mol/kg$= 3 mol/kg

So $\Delta T_{f}=(1)\times (1.86)\times (3)^{0}C$ = -5.58$^{0}C$

Answer 2

is this for a test or are you genuinely interested? molality = mols sugar/kg solvent

Solve for molality

delta T = Kf*m

Solve for delta T and subtract from zero C to find the new freezing point.

or

-5.58