An aqueous solution of sodium bromide (MW = 102.9 g/mol) freezes at
-1.09 oC. What is the total molality of solute particles? Based on
this information, how many grams of sodium bromide would be present in
one kg solvent? Assume an ideal value for the van't Hoff factor.

So to answer your homework question check out the following link.  It
will demonstrate the calculation you will need to perform.

http://www.nyu.edu/classes/tuckerman/honors.chem/lectures/lecture_13/node9.html

If you have any more questions take alook at the links from the
following google search -

://www.google.com/search?hl=en&ie=ISO-8859-1&q=freezing+point+depression&btnG=Google+Search

Let "x" grams of the solute (Sodium Bromide) be dissolved in 1 Kg of water. 
Thus, the molal concentration, m = x/102.9 molal.

Now, apply the equation, dT = -iKm, wher dT is the freezing
depression, i is the vant Hoff factor, K is the freezing point
depression constant and m is the molal concentration. In our case, dT
= -1.09 0C, i = 2 (since the Sodium Bromide contains 2 ions), K =
-1.86 (constant for water).

Thus, (-1.09) = -(2) (-1.86) (x/102.9)
=> x = 30.15 grams. This answers the second question.

The molality of the solute particles is the number of moles of solute
in a kg of solvent. Thus, the molaity in our case is (30.15/102.9) =
0.29 molal. This answers the first question.