Question

What would the final freezing point of water be if 3 mol of

sugar were added to 1 kg of water (Kx = 1.86°C/(mol/kg)
for water and i = 1 for sugar)?
A. -1.86°C
B. -5.58°C
C. -0.62°C
Ο Ο
D. +5.58°C
SUBMIT

Answer 1

B. -5.58°C.

Step-by-step explanation:

Hello!

In this case, since the freezing point depression is computed as follows:

`\Delta T_f=-i*m*Kf`

Whereas i=1 as the van't Hoff's factor of sugar (nonionizing solute), m=3mol/1kg=3mol/kg as the molality and Kf=1.86 °C/(mol/kg) as the freezing point depression constant for water. In such a way, we plug in to obtain:

`\Delta T_f=-1*3(mol)/(kg) *1.86(\°C)/(mol/kg) \\\\\Delta T_f=-5.58\°C`

Now, since the freezing point of pure water is 0°C, we infer that freezing point of such solution is:

B. -5.58°C.

Best regards!