Question

A solution contains 20.0 g of C6H12O6 (glucose) in 250 g. of water.

A. what is freezing point depression of solvent?

Answer 1

Molar mass glucose = 180 g/mol

number of moles:

20 / 180 = 0.111 moles

Δ t f * ( Kf) *( i ) * ( m )

250 g of water in Kg = 250 / 1000 = 0.25 kg

1.86 ºC/m * 0.111 / 0.25

1.86ºC/m * 0.444

= 0.82584ºC

Fp = 0ºC - 0.82584

Fp = - 0.82584ºC

hope this helps!

Answer 2

The freezing point depression of solvent is 0.8265 K.

Step-by-step explanation:

Mass of glucose = 250 g

Mass of solvent = 20.0 g = 0.020 kg

Molal depression water constant of water,
`K_f` = 1.86 K kg/mol

Molality of the solution =
`(20 g)/(180 g/mol* 0.250 kg)=0.4444 mol/kg`

Normal freezing point of water =
`T=273 K`

Freezing point of solution =
`T_f`

`\Delta T_f=T-T_f`

`\Delta T_f=K_f* molality=1.86 K kg/mol * 0.4444 mol/kg=0.8265 K`

The freezing point depression of solvent is 0.8265 K.