Question

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

Answer 1

Answer:

The freezing point will be -2.046°C.

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;

$\Delta T_f = 1.86 \times 1.1 \times 1 = 2.046 ^{\circ}C$

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