**1. Introduction**

**1.1 Background Research (Ain & Yee Theng)**

Freezing point depression is a colligative property, a property that depends on the number of solute particles are in the solvent. The van't Hoff factor on the other hand, is how a molecule of a solute dissociates, or breaks apart, in the solvent. Covalent compounds, like sucrose (C

_{12}H_{22}O_{11}), do not dissociate in solution. These compounds have van't Hoff factors i = 1. Ionic compounds, like table salt (NaCl), dissociate when in solution.( Eli, Todd & Keith. (n.d.)) Table salt (NaCl) has a van't Hoff factor i = 2 because it dissociates into two ions in solution Na^{+}and Cl^{-}. The third factor, the molal freezing-point-depression constant, K_{f}, is different for every solvent. It has units of (° C/m), and it tells indicates how much 1 mol of solute added to 1 kg of solvent will lower the solvent's freezing point. For pure water, K_{f}= 1.86° C/m. (Lachish, U. (2000)) Combining these three factors—molality, m, van't Hoff factor, i, and molal freezing-point-depression constant, K_{f}—into an equation that predicts how much the freezing point of a solvent will decrease, ΔT, when a certain amount of solute is added. (J.B. Condon, Roana State Community College. (n.d.))
The freezing point depression formula is defined as shown below:

Degrees Freezing Point is Depressed (° C) = Molal Freezing-Point-Depression Constant (° C/m)

× molality of solution (mol solute/kg solvent) × van't Hoff Factor (J.B. Condon, Roana State Community College. (n.d.))

In terms of symbols, the freezing point depression formula is stated below with what the symbols stands for:

ΔT = K

_{f}m i
ΔT - freezing point in degrees Celsius (° C)

K

_{f - }molal freezing point depression constant in degrees Celsius per molal (° C/m)
m - molality of the solution in moles per kilogram (mol/kg)

i - van't Hoff factor of the solute

The equation below shows the formula for the freezing point of a solution:

Solution Freezing Point (° C) = Solvent Freezing Point (° C) - Degrees Freezing Point is Depressed (° C)

In terms of symbols, the freezing point of a solution is as stated below:

T

_{n}= T_{f}- ΔT
T

_{n }- freezing point of the solution in degrees Celsius (° C)
T

_{f}- freezing point of the solvent in degrees Celsius (° C)
ΔT i- freezing point depression in degrees Celsius (° C)

(Science Buddies Organization. (n.d.))

An example of how the equation works is as stated below:

Calculate the freezing point of a solution of 5.00 g of diphenyl C

_{12}H_{10}and 7.50 g of naphthalene, C_{10}H_{8}dissolved in 200.0 g of benzene (fp = 5.5 °C)
Solution

There is a tiny curve in this problem, but keep in mind that colligative properties are all about how many particles in solution and nothing else.

1) The key to this problem is to calculate moles of each substance and add then together:

(5.00 g / 154.2 g mol¯

^{1}) + (7.50 g / 128.2) = 0.0909 mol
2) Calculate the molality:

0.0909 mol / 0.200 kg = 0.455 m

3) Calculate the freezing point depression:

ΔT = i K

_{f}m
x = (1) (5.12 °C m¯

^{1}) (0.455 m) = 2.33 °C
The solution freezes at 5.5 - 2.33 = 3.17 °C

**1.2 Research Question**

Does the amount of chemical dissolved in water increases the freezing point of water?

**1.3 Hypothesis**

The higher amount of dissolving chemicals in water, the lower the freezing point of water.

**1.3.1 Independent variable(s)**

Amount of chemical added (g)

**1.3.2 Dependent variable**

The freezing temperature of chemical solution (℃)

**1.3.3 Constants**

Amount of water added, the duration of experiment, the surrounding room temperature, the type of water, the type of chemicals used and the type of water used.

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