Answer:
0.0078 L of the stock solution is required to make up 1.00 L of 0.13 M [tex]HNO_{3}[/tex]
Explanation:
According to laws of equivalence, [tex]C_{1}V_{1}=C_{2}V_{2}[/tex]
where, [tex]C_{1}[/tex] and [tex]C_{2}[/tex] are initial and final concentration respectively. [tex]V_{1}[/tex] and [tex]V_{2}[/tex] are initial and final volume respectively.
Here, [tex]C_{1}=16.6M[/tex], [tex]C_{2}=0.13M[/tex] and [tex]V_{2}=1.00L[/tex]
So, [tex]V_{1}=\frac{C_{2}V_{2}}{C_{1}}[/tex]
or, [tex]V_{1}=\frac{(0.13M)\times (1.00L)}{(16.6M)}[/tex]
or, [tex]V_{1}=0.0078L[/tex]
Hence 0.0078 L of the stock solution is required to make up 1.00 L of 0.13 M [tex]HNO_{3}[/tex]
