Molarity vs. molality - 4 | Chemical Calculations

mass% = $\frac{\mathrm{mass} \mathrm{solute}}{\mathrm{mass} \mathrm{solution}}$ x 100

⇒ in the solution, there are 30.0 g of NaOH in 100 g of solution

Step 1: determine the mass of NaOH m_NaOH in one liter of solution:

m_NaOH (in g) = V (in mL) x d (in g.mL^-1) x $\frac{\mathrm{mass}\%}{100}$ = 1000 x 1.325 x 0.300 = 397.5 g
 

Step 2: determine the molarity of the solution:

n_NaOH = $\frac{{\mathrm{m}}_{\mathrm{NaOH}}}{{\mathrm{M}}_{\mathrm{NaOH}}}$    and    n_NaOH = M_sol x V_sol

⇒ M_sol x V_sol = $\frac{{\mathrm{m}}_{\mathrm{NaOH}}}{{\mathrm{M}}_{\mathrm{NaOH}}}$

⇒ M_sol (in mol.L^-1) = $\frac{{\mathrm{m}}_{\mathrm{NaOH}} (\mathrm{in} \mathrm{g})}{{\mathrm{M}}_{\mathrm{NaOH}} (\mathrm{in} \mathrm{g}.{\mathrm{mol}}^{-1}) \mathrm{x} {\mathrm{V}}_{\mathrm{sol}} (\mathrm{in} \mathrm{L})}$ = $\frac{397.5}{40.0 \times 1.00}$ = 9.94 mol.L^-1