Calculate the molarity of a solution that contains 0.012 moles of solute and 23.1 grams of solvent.

We can use the balanced chemical equation between oxalic acid and sodium hydroxide to determine the amount of oxalic acid present in the solution: {eq}H_2C_2O_4(aq) + 2NaOH(aq) rightarrow Na_2C_2O_4(aq) + 2H_2O(l) {/eq} From the equation, we know that 1 mole of oxalic acid reacts with 2 moles of NaOH. Therefore, we can calculate the moles of NaOH used in the titration: moles of NaOH = (0.1573 M) x (25.80 mL/1000 mL) = 0.00405 moles Since 1 mole of oxalic acid reacts with 2 moles of NaOH, the moles of oxalic acid in the 10.00 mL sample is: moles of H2C2O4 = 0.00405 moles / 2 = 0.002025 moles The molarity of the oxalic acid solution is then: Molarity = moles of H2C2O4 / volume of solution (in L) = 0.002025 moles / 0.01000 L = 0.2025 M Therefore, the molarity of the oxalic acid solution is 0.2025 M.