If you combine 23.0 g of a solute that has a molar mass of 23.0 g/mol with 100.0 g of a solvent, what is the molality of the resulting solution?

10 M  
Technically, this is a bad question and the actual answer is impossible to determine. The reason for this is that molarity is DEFINED as moles per liter. There's no indication of the volume that 100.0 grams of solvent requires and hence there's no sure way of determining the molarity of resulting solution. But for this problem I will assume that the solvent is water and that the density of the solvent is 1 gram per milliliter and further, I will assume that after dissolving the solute into the solvent, the volume of the solvent doesn't change. So let's determine how many moles of solute we have:
 23.0 g / 23.0 g/mol = 1 mol  
So we have 1 mole of solute. And the volume of solvent is 100 ml or 0.100 liters. So 
1 mol / 0.100 L = 10 mol/L = 10 M  
As a side note. The usual method of producing a solution of known molarity is the dissolve the solute in a smaller quantity of solvent. Then after all of the solute has been dissolved, add more solvent until the desired volume of solution is reached. That compensates for the fact that the volume of the solution changes as solute is added. For instance, when adding sodium chloride to water, the volume actually DECREASES at lower concentrations before it finally starts to increase at higher concentrations of sodium chloride.