What is the freezing point of an aqueous 1.00 m NaCl solution? (Kf = 1.86 *C/m) (Assume complete dissociation of the salt.)

The freezing point is -3.72 degrees Celsius

Explanation:

In this question, by assuming the complete dissociation of the salt, we want to calculate the freezing point of an aqueous 1 M NaCl solution .

We proceed as follows;

Firstly, we write the dissociation equation;

NaCl ——> Na+ + Cl-

So the number of ions present in the dissociation is 2. This means that the Vant Hogg factor also called I = 2

Mathematically;

ΔTf = Kf*m*i

m represents the molality which is given as 1

So by substituting the values, we have;

ΔTf = 1.86 * 1 * 2 = 3.72 degrees Celcius

The freezing point however is 0 - ΔTf = 0 - 3.72 = -3.72 degrees Celcius

question: ray weighs 100 kilograms, his wrestling partner, bobby, weighs 50 kilograms. if ray moves at 4 m s , how fast will bobby have to run at him to apply an equal amount of kinetic energy?

answer: 8 m/s he needs to run twice as fast to keep up so he has some work to do

explanation:   if the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.

question answered by

(jacemorris04)