# What is the freezing point of a solution made by mixing 30.0 grams of NH3 with 300.0 grams of water (Kf for water is 1.86 degrees celsius per molal)?

## 1. Solve for the Molality of NH3(aq)

Molality (molal) is in units of moles of solute per kilograms of solvent. Ammonia is the solute and water is the solvent.

### 1a. Calculate Moles Solute (NH3) moles NH3 = (30.0 g NH3)/1 x (1 mol NH3)/(17.04 g NH3) = 1.76 mol NH3

### 1b. Calculate Kilograms Solvent (H2O) kg H2O = (300.0 g H2O)/1 x (1 kg H2O)/(1000 g H2O) = 0.3000 kg H2O

### 1c. Calculate Molality NH3(aq) molal NH3(aq) = (1.76 mol NH3)/(0.3000 kg H2O) = 5.87 mol NH3/kg H2O = 5.87 molal NH3(aq)

## 2. Solve for the Freezing Point of the Solution [NH3(aq)]

### 2a. Calculate the Decrease in Freezing Point

The decrease in freezing point is equal to the van’t Hoff factor multiplied by the molal freezing-point depression constant for water and the molality of ammonia solute particles in water. Decrease in freezing point = (1)(1.86 oC/molal)(5.87 molal) = 10.9 oC

### 2b. Calculate the Freezing Point of the Solution

The freezing point of the solution is equal to the difference between the normal freezing point of pure water and the decrease in freezing point value. Freezing point of the solution = (0 oC) – (10.9 oC) = -11 oC