Dissolving 120g of urea (mol.wt 60) in 1000g of water gave a solution of density 1.15 g/mL. What is the molarity of the solution?

The idea here is that a solution's molarity tells you the number of moles of solute present in #"1 L"# of solution.

So in order to calculate a solution's molarity, you essentially need to know the number of moles of solute present in exactly #"1 L" = 10^3# #"mL"# of solution.

Start by using the molar mass of urea to calculate the number of moles present in your sample

#120 color(red)(cancel(color(black)("g"))) * "1 mole urea"/(60color(red)(cancel(color(black)("g")))) = "2 moles urea"#

Now, you know that your solution contains #"120 g"# of urea, the solute, and #"1000 g"# of water, the solvent. This implies that the total mass of the solution

#"mass solution = mass solute + mass solvent"#

will be equal to

#"mass solution" = "120 g + 1000 g" = "1120 g"#

You also know that this solution has a density of #"1.15 g mL"^(-1)#, which means that every #"1 mL"# of solution has a mass of #"1.15 g"#.

Use the density of the solution to calculate its volume

#1120 color(red)(cancel(color(black)("g"))) * "1 mL"/(1.15color(red)(cancel(color(black)("g"))) ) = "973.9 mL"#

Now, your goal is to figure out the number of moles of solute present in #10^3# #"mL"# of solution, so use the known composition of the solution as a conversion factor to get

#10^3 color(red)(cancel(color(black)("mL solution"))) * "2 moles urea"/(973.9color(red)(cancel(color(black)("mL solution")))) = "2.0536 moles urea"#

You can thus say that the molarity of the solution is equal to

#color(darkgreen)(ul(color(black)("molarity = 2.1 mol L"^(-1))))#

I'll leave the answer rounded to two sig figs.