Question

I can't solve it T.T please help meh what is the freezing point and boiling point of 50% by mass/weight of sugar syrup C12H22O11?

Answer 1

`"Freezing point = -5.43 °C"`

`"Boiling point = 101.5°C"`

Explanation:

Molar mass of `"C"_12"H"_22"O"_11 = "342 g/mol"`

Let mass of solution be `100` grams. Then, mass of solute will be `50` grams and mass of solvent will also be `50` grams (`∵` it is given that solution is `50%` mass by weight)

`"Moles of solute" = "50 g"/"342 g/mol" = "0.146 mol"`

`"Molality" = "Moles of solute" / "Mass of solvent (in kg)" = "0.146 mol"/"0.05 kg" = "2.92 molal"`

Depression in freezing point point is given as

`DeltaT_f = -K_fm`

Where

• `K_f` is the freezing point constant of solvent, which is `1.86^@C"/"m` for water

• `m` is the molality of solution

`(T - 0°C) = -"1.86 °C"/cancel"molal" × 2.92cancel"molal"`

`T = -5.43 °C`

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Elevation in boiling point is given as

`DeltaT_b = K_bm`

Where

• `K_b` = Freezing point constant of solvent = 0.52 °C/m for water

• `m` = molality of solution

`(T - 100°C) = "0.52 °C"/cancel"molal" × 2.92cancel"molal"`

`T = 101.5°C`