Question

Calculate the freezing point of a 2.6-molal aqueous sucrose solution. The freezing point depression constant for water is 1.86 degrees C/molal.

A.) 4.8 °C
B.) 1.4 °C
C.) -1.4 °C
D.) -4.8 °C

Answer 1

Freezing point formula is Molarity * depression constant. And for this specific problem, you should get 4.8 as the result but since the freezing point is always negative, then the answer should be -4.8. The correct answer between all the choices given is the last choice or letter D. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

Answer 2

Answer: D.) -4.8 °C

Explanation:

The formula used for depression in freezing point is:

$\Delta T_f=K_f\times m$

$\Delta T_f$ = change in freezing point

$K_f$ = freezing point constant = $1.86^0C/m$

$\Delta T_f=T_f^0-T_f$

$\Delta T_f=1.86^0C/m\times 2.6m$

$\Delta T_f=4.836^0C$

$T_f^0-T_f=4.836^0C$

$0^C-T_f=4.836^0C$

$T_f=-4.8^0C$

Thus the freezing point of the solution is $-4.8^0C$.