Answer to Question #155091 in General Chemistry for Saskia

Question

Calculate the volume of stock solution nitric acid of concentration 4.00 M which is required to prepare 60.0 mL of nitric acid of concentration 0.200 M. 

Answer 1

$c_1 = 4.00M\\ v_1 = \ ?\\ c_2 = 0.200M\\ v_2 = 60.0mL$

Using the formula,

$c_1v_1 = c_{2}v_2$

$4 × v_1 = 0.200 × 60.0$

$v_{1} = \dfrac{0.200×60.0}{4}$

$v_1 = 3mL$

$\therefore$The volume of stock solution of nitric acid required is 3 mL.