$$M_{HNO3} = \frac{M_{NaOH} \times V_{NaOH}}{V_{HNO3}}$$
where:
- \(M_{HNO3}\) is the molarity of the HNO3 solution
- \(M_{NaOH}\) is the molarity of the NaOH solution (given as 1.67 M)
- \(V_{NaOH}\) is the volume of the NaOH solution used (given as 10.0 mL)
- \(V_{HNO3}\) is the volume of the HNO3 solution used (20.0 mL)
Substituting the given values, we get:
$$M_{HNO3} = \frac{1.67 \text{ M} \times 10.0 \text{ mL}}{20.0 \text{ mL}} = 0.835 \text{ M}$$
Therefore, the molarity of the HNO3 solution is 0.835 M.
