Question

If the molarity of sugar is 1.1, what will the freezing point be?

Answer 1

The freezing point will be -2.046°C.

Step-by-step explanation:

The freezing point depression equation is
`\Delta T_f = k_f \cdot m \cdot i`

Where;

`\Delta T_f` = The temperature depression of the freezing point

`k_f`= The constant of freezing point depression which is solvent dependent = 1.86°C/m

i = The number of particles the substance decomposes into in solution = 1 for sugar (a covalent compound)

m = The molality of the solution = 1.1

Therefore, we have;

`\Delta T_f = 1.86 * 1.1 * 1 = 2.046 ^(\circ)C`

Therefore the freezing point will be 0 - 2.046°C = -2.046°C.