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

Answer:

B. -5.58°C.

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:

$\Delta T_f=-1*3\frac{mol}{kg} *1.86\frac{\°C}{mol/kg} \\\\\Delta T_f=-5.58\°C$

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!