Asodium nitrate solution is 21.5% (by mass) of nano3 (molar mass = 85.00 g/mol) and the solution has a density of 1.08 g/ml. calculate the molarity (m) of the solution.

Density of the solution is 1.08 g/ml.
That means, mass of 1 ml = 1.08 g

Let's consider 1000 ml ( 1 L) of solution.
Then mass of 1 L solution = 1.08 g/ml x 1000 ml
                                          =  1.08 x 10³ g

The mass percentage NaNO₃ is 21.5% means 21.5 g of NaNO₃ in 100 g of solution.
Hence, the mass of NaNO₃ in 1 L of solution = 1.08 x 10³ g x 21.5 / 100
                                                                        = 232.2 g
Moles = mass / molar mass

Hence, moles of NaNO₃ in 1 L of solution  = 232.2 g / 85.00 g/mol
                                                                    = 2.73 mol

Molarity = moles of solute (mol) / Volume of the solution (L)

Hence molarity of NaNO₃ =  2.73 mol / 1 L
                                            = 2.73 mol L⁻¹

Answer is: molarity of sodium nitrate solution is 2.73 M.
V(NaNO₃) = 1 L · 1000 mL/L = 1000 mL.
d(NaNO₃) = 1.08 g/mL.
ω(NaNO₃) = 21.5% ÷ 100% = 0.215.
mr(NaNO₃) = V(NaNO₃) · d(NaNO₃).
mr(NaNO₃) = 1000 mL · 1.08 g/mL.
mr(NaNO₃) = 1080 g.
m(NaNO₃) = ω(NaNO₃) · mr(NaNO₃).
m(NaNO₃) = 0.215 · 1080 g.
m(NaNO₃) = 232.2 g.
n(NaNO₃) = m(NaNO₃) ÷ M(NaNO₃).
n(NaNO₃) = 232.2 g ÷ 85.00 g/mol.
n(NaNO₃) = 2.73 mol.
c(NaNO₃) = n(NaNO₃) ÷ V(NaNO₃).
c(NaNO₃) = 2.73 mol ÷ 1 L.
c(NaNO₃) = 2.73 mol/L.

And a) oxygen i believe