I have two questions for a pre-lab in chemistry.... The first one is: 1. What is

ID: 794127 • Letter: I

Question

I have two questions for a pre-lab in chemistry....

The first one is:

1. What is the molality of a solution consisting of 16.035 g of NaCl dissolved in 100.017 g of water?

The second one is:

2. What would be the expected freezing point of a solution of 16.420 g NaCl dissolved in 100.059 g of water?

I believe you use the formula: triangleTf=Kf*m*n

where triangle Tf is the observed (degrees Celcius) freezing point depression of the solution;

Kf is the molal freezing point constant (characteristic of the solvent, 1.860 degress Celcius/m for water);

m is the molality of the solution; and

n is the number of particles formed from the dissociation of each formula unitof solute.

1)molality = wt/M.wt (1000/Vol. in ml)

Molality = 16.035/58.5(1000/100.017) = 2.714 mol/Kg

2) triangleTf=Kf*m*n

0 -Tf = 1.860*(16.42/58.5*1000/100.059)*2

Tf = - 10.43 ^0c