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

D.) -4.8 °C

Explanation:

The formula used for depression in freezing point is:

= change in freezing point

= freezing point constant =

Thus the freezing point of the solution is .

the answer to your question is c.

if you read the house of usher you should know that the house was haunted

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%