Answer: Molar concentration of nitric acid in the final solution is [tex]8.12\times 10^{-3}M[/tex]
Explanation:
According to the dilution law,
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1[/tex] = molarity of stock [tex]HNO_3[/tex] solution = 0.1015 M
[tex]V_1[/tex] = volume of stock [tex]HNO_3[/tex]solution = 200 ml
[tex]M_2[/tex] = molarity of dilute [tex]HNO_3[/tex] solution = ?
[tex]V_2[/tex] = volume of dilute [tex]HNO_3[/tex] solution = (2300 +200 )ml = 2500 ml
Putting in the values we get:
[tex]0.1015M\times 200=M_2\times 2500[/tex]
[tex]M_2=8.12\times 10^{-3}M[/tex]
Thus the molar concentration of nitric acid in the final solution is [tex]8.12\times 10^{-3}M[/tex]
