0.53L
According to dilution formula:
[tex]C_1V_1=C_2V_2[/tex]C1 and C2 are the initial and final concentrations
V1 and V2 is initial and final volume
Given the following parameters
• V1 = 45mL
,• C2 = 0.2M
,• V2 = 119.4mL
Substitute the given parameters into the formula
[tex]\begin{gathered} C_1=\frac{C_2V_2}{V_1} \\ C_1=\frac{0.2\times119.4}{45} \\ C_1=\frac{23.88}{45} \\ C_1=0.53M \end{gathered}[/tex]Hence the molarity of the nitric acid is 0.53L
