# Freezing point

1. Nov 28, 2004

### parwana

How many grams of urea (MW = 60.056 g/mol) would have to be dissolved in 115.0 grams of water to lower the freezing point by 1.60 degrees celsius?

Tf= Kf x Cm

Kf= 1.858
Cm= molality of solute in kg

2. Nov 28, 2004

### chem_tr

Here, $T_f$ is 1.60, and from there you can calculate the molality:

$$T_f=K_f \times C_m and C_m=\frac {T_f}{K_f}$$

$$C_m=\frac {1.60}{1.858}=0.861$$

Since your molality is 0.861, it means that in 1 kg of solution you'll have to dissolve moles of this amount of compound.

3. Nov 28, 2004