The freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is - 0.45°C
Using the equation,
Δ
= i
m
where:
Δ = change in freezing point (unknown)
i = Van't Hoff factor
= freezing point depression constant
m = molal concentration of the solution
Molality is expressed as the number of moles of the solute per kilogram of the solvent.
Molal concentration is as follows;
MM KCl = 74.55 g/mol
molal concentration = 
molal concentration = 0.1219m
Now, putting in the values to the equtaion Δ = im we get,
Δ = 2 × 1.86 × 0.1219
Δ = 0.4536°C
So, Δ of solution is,
Δ
= 0.00°C - 0.45°C
Δ = - 0.45°C
Therefore,freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is - 0.45°C
Learn more about freezing point here;
brainly.com/question/3121416
#SPJ4
Answer:
Nov 28, 2019 - These templates were designed to help get you started. ... of each opportunity is clear to see, because the ones you send match not only my interests but my abilities. What you do is really motivating and keeps me uplifted in my job ... to thank you for all the support you've shown me throughout my career, ...
Explanation:
The nuclear equation that describes the alpha decay of Polonium-210 can be written like this:
210/
84Po→206/82Pb+4/2 He
Answer:
copper
Explanation:
it's on the periodic table of elements
