A solution is prepared with 50.0 g of HNO3 and 130.0 g of H2O. It has a density of 1.21 g/mL. What is its molarity (M)?

%by mass = (50 g HNO3/180 g soln)*100 = 27.777% which I would keep although that is too many significant figures (s.f.). (I would round at the end.).

density = 1.21 g/mL; therefore, 1000 mL has a mass of 1.21 g/mL x 1000 mL = 1210 grams.
That is 27.777% HNO3; therefore, g HNO3 there is 1210 x 0.27777 = 336.1 grams HNO3.
How many moles is that?
moles = grams/molar mass = ??
That many moles in 1 L soln is that molarity. Round your answer to 3 s.f.

To find the molarity (M) of the solution, we need to know the moles of solute (HNO3) and the volume of the solution.

First, we need to calculate the volume of the solution. We know the density of the solution is 1.21 g/mL, and we have 130.0 g of water added to it. We can calculate the volume using the formula:

Volume (mL) = Mass (g) / Density (g/mL)

Volume (mL) = 130.0 g / 1.21 g/mL
Volume (mL) ≈ 107.44 mL

Next, we need to calculate the moles of HNO3 in the solution. We know the mass of HNO3 is 50.0 g. We can calculate the moles using the formula:

Moles = Mass / Molar mass

The molar mass of HNO3 is:
H = 1.01 g/mol
N = 14.01 g/mol
O = 16.00 g/mol (x3)

Molar mass of HNO3 = (1.01 g/mol) + (14.01 g/mol) + (16.00 g/mol x 3)
Molar mass of HNO3 = 63.01 g/mol

Moles of HNO3 = 50.0 g / 63.01 g/mol
Moles of HNO3 ≈ 0.793 mol

Finally, we can calculate the molarity (M) of the solution using the formula:

Molarity (M) = Moles / Volume (L)

Volume (L) = 107.44 mL / 1000 mL/L
Volume (L) = 0.10744 L

Molarity (M) = 0.793 mol / 0.10744 L
Molarity (M) ≈ 7.37 M

Therefore, the molarity of the solution is approximately 7.37 M.