Answer:
0.3634 °C
Step-by-step explanation:
In this question, we are tasked with calculating the freezing point depression of 0.19 M C12H12O11( sucrose)
Mathematically this can be calculated using the formula;
ΔTf = i * kf * m
where ΔTf is the freezing point depression
i is Van’t Hoff factor
kf is the molal freezing point depression for the solvent
m is the molality of the solution
From the question we identify the following;
Since sucrose is a non-electrolyte, i = 1 ( i refers to the number of moles of particles obtained when 1 mol of solute dissolves)
m is 0.19
kf = 1.86
substituting these values into the equation, we have
ΔTf = 1 * 1.86 * 0.19 = 0.3534 °C