We're given molarity (indirectly), and density.

So, let's find the molarity,

#(230g)/L * (mol)/(58.5g) approx 3.93M#

If there are #230g# solute and 1148g total in a liter of solution, then

#(230g)/(1148g) * 100% approx 20.03%#

of the solution is solute and by implication, #79.97%# of the solution is solvent.

The volumetric percent of water is very close to #100%#; we're never told any other solutions are "combined".

We already deduced #3.93mol# of solute are in the solution, and,

#1148g - 230g = 918g * (mol)/(18g) approx 51.0mol#

of solvent are in the solution. Hence,

#(3.93mol)/(3.93mol+51.0mol) approx 0.0715#

is the mole fraction of sodium chloride, and by implication #0.928# is the mole fraction of water.

We already deduced #918g * (kg)/(10^3g) = 0.918kg# of solvent are in the solution, and #3.93mol# of solute are in the solution, hence,

#(3.93mol)/(0.918kg) approx 4.28m#

Finally, the last bit is a simple unit conversion,

#230g * (kg)/(10^3g) * (2.20lb)/(kg) approx 0.506lbs#

of solute are distributed within the solution.