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

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

Answer 2

The final freezing point of water is -5.58
`^(0)C`

Step-by-step 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 =
`(number of moles of sugar)/(mass of water in kg)` =
`(3)/(1)mol/kg`= 3 mol/kg

So
`\Delta T_(f)=(1)* (1.86)* (3)^(0)C` = -5.58
`^(0)C`