Someone fr ! what is the freezing point in celsius of a 1.56 m aqueous solution of cacl2? (report amount to 3 decimal points)

-8.705 °C (amount to three decimal points/ significant figures)

Explanation:

1. This is a water solution so, you will need to know the freezing and boiling point of water.

Freezing point of water in Kf (°C/m) = 1.86 °C/m

Boiling point of water in Kb (°C/m) = 0.512 °C/m

2. There are 3 number of particles in CaCl2

3. Multiple the number of particles and the 1.56 m to the freezing and boiling point of water.

4. (Freezing Point) Kfm (1.86)*(1.56)*(3) = 8.7048

5. (Boiling Point) Kbm (0.512)*(1.56)*(3) = 2.39616

6. Subtract the Freezing point of water in °C (Celsius) which is 0.0 °C from the answer 8.7048 ( 0 - 8.7048 = -8.7048 °C

7. Add the Boiling point of water in °C (Celsius) which is 100.0 °C from the answer 2.39616 ( 100 + 2.39616 = 102.39616)

8. The question is asking for the Freezing point of 1.56 m which is what we now know to be -8.7048 °C

9. The answer in 3 decimal points will round up to -8.705 °C

10. The reason is because 0 does not count as a significant figure so you need the “5” to make this answer have 3 significant figures which are the numbers 8, 7 and 5 which makes it round up to the 3 decimal points.