Question

How many mL of a stock solution of 9.0 A/HNO_3 would you have to use to prepare 0.380 L of 0.450 M HNO_3?

Answer 1

So the amount of mols of HNO3 in the tarjet solution can be calculated as the Concentration times the volume as the following formula shows:

$m=[C_{HNO_{3}}]\cdot V$

and those mols have to be tha same as the stock solution used so we can do the following

$m=[0.450\frac{mol}{L}]\cdot 0.380L$

$0.171 mol=[9\frac{mol}{L}]\cdot V$

$V=\frac{0.171mol}{9\frac{mol}{L}}$

$V=0.019L$

$V=0.019L\cdot \frac{1000 ml}{1L}=19ml$