# Question b886a

May 24, 2017

Here we have a dilution chemistry problem. For problems like this one, we use the following:

color(white)(aaaaaaaaaaaaaa)color(purple)(M_1V_1 = M_2V_2

Where
${M}_{1} = \text{molarity of solution 1}$ $\left(3.0 \text{ M } H N {O}_{3}\right)$
${V}_{1} = \text{volume of solution 1 (??)}$
${M}_{2} = \text{molarity of solution 2}$ $\left(0.250 \text{ M } H N {O}_{3}\right)$
${V}_{2} = \text{volume of solution 2 (0.040 L)}$

We know that molarity is defined as moles of solute per liter of solution.

color(white)(aaaaaaaaa)"Molarity" = "moles of solute"/(1" Liter solution")

The question

$\textcolor{w h i t e}{- - - -}$How would you prepare 40.0 mL of 0.250 M HNO3 from a stock (initial) solution of 3.00 M HNO3

is just asking you how would one make a diluted solution of $H N {O}_{3}$ from a concentrated stock solution of $H N {O}_{3}$.

Steps

• ${M}_{1} {V}_{1} = {M}_{2} {V}_{2}$

• $\left(3.0 \text{ M")(V_(1)) = (0.250" M")(0.040" L}\right)$

• V_(1) = [(0.250cancel" M")(0.040" L")]/(3.0cancel" M")#

• ${V}_{1} = 0.0033 \text{ L" or 3.3" mL}$

This means you would need $3 \text{ mL}$ of the concentrated stock solution of $H N {O}_{3}$ and then you would need to dilute it down with water until you reach the $40 \text{ mL}$ mark.