Verified Answer :

Willchemistry

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:

The freezing point depression of solvent is 0.8265 K.

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 = \(\frac{20 g}{180 g/mol\times 0.250 kg}=0.4444 mol/kg\)

Normal freezing point of water = \(T=273 K\)

Freezing point of solution = \(T_f\)

\(\Delta T_f=T-T_f\)

\(\Delta T_f=K_f\times molality=1.86 K kg/mol \times 0.4444 mol/kg=0.8265 K\)

The freezing point depression of solvent is 0.8265 K.

Related Questions