How to calculate molality of a saline solution that contains 175.5 g of NaCl / L

[farisallil]

A saline solution contains 175.5 g of NaCl per liter of solution. Using a water density of 0.9982 g/cm^3, express the concentration of NaCl in term of molality.

I know that molality = moles of solute/ # of kilograms of solvent. But how do I get the number of kilograms of solvent from this problem?

[kingchemist]

Mass of the solution (water + salt) = Density x Volume = 0.9982 x 1000 = 998.2 g

Mass of water = (998.2 - 175.5)g

I think you should be OK from here

[farisallil]

Mass of the solution (water + salt) = Density x Volume = 0.9982 x 1000 = 998.2 g

Mass of water = (998.2 - 175.5)g

I think you should be OK from here

Well, 0.9882 is the density of pure water not the solution

[kingchemist]

OK, I misread the question. However, you will need to be given the density of the solution to work this out.

[farisallil]

ok
I have looked into the textbook and found the attached table for densities of NaCl solutions
Here is my try after I found the table:
175.5 g/L = 1.4646 lb/gal
(1.587-1.356)/(1.1162-1.1009)=(1.587-1.464)/(1.1162-x)
--> specific gravity = x= 1.10809
specific gravity * density of water = density of solution
--> 1.10809*0.9982 = 1.1061 g/cm^3

density of solution * volume = 1.1061*1000 = 1106.1
mass of water = 1106.1-175.5=930.6 grams=0.9306 kg

molality = (175.5/58.43)/(0.9306)= 3.228 molal

However, the textbook says that the final answer is 3.198
so what did I do wrong?