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

Answer: D.) -4.8 °C

Step-by-step explanation:

The formula used for depression in freezing point is:

`\Delta T_f=K_f* 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* 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`.

Answer 2

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.