Answer:
[tex]\large \boxed{\text{1.4 mol/L}}[/tex]
Explanation:
(a) Balanced equation
2NaOH + H₂SO₄ ⟶ Na₂SO₄ + 2H₂O
(b) Moles of H₂SO₄
[tex]\text{Moles of H$_{2}$SO}_{4} = \text{32.5 mL H$_{2}$SO}_{4} \times \dfrac{\text{1.0 mmol H$_{2}$SO$_{4}$}}{\text{1 mL H$_{2}$SO$_{4}$}}\\= \text{32.5 mmol H$_{2}$SO}_{4}[/tex]
(c) Moles of NaOH
The molar ratio is 2 mol NaOH:1 mol H₂SO₄.
[tex]\text{Moles of NaOH} = \text{32.5 mmol H$_{2}$SO}_{4} \times\dfrac{\text{2 mmol NaOH}}{\text{1 mmol H$_{2}$SO}_{4}}\\\\= \text{65 mmol NaOH}[/tex]
(d) Molar concentration of NaOH
[tex]c = \dfrac{\text{moles of solute}}{\text{litres of solution}}\\\\c = \dfrac{n}{V}\\\\c= \dfrac{\text{65 mmol}}{\text{45.0 mL}} = \text{1.4 mol$\cdot$L$^{-1}$}\\\\\text{The molar concentration of the NaOH is $\large \boxed{\textbf{1.4 mol/L}}$}[/tex]
Note: The answer can have only two significant figures because that is all you gave for the molar concentration of the sulfuric acid.
