# Question 77da8

Dec 9, 2014

$\text{47.8 mL}$

#### Explanation:

A solution's concentration percentage by mass is defined as the mass of the solute divided by the total mass of the solution and multiplied by 100.

Nitric acid's concentration by mass is given to be 70.3%; this means that for every $\text{100 g}$ of solution we get $\text{70.3 g}$ of ${\text{HNO}}_{3}$.

Let's assume we have a $\text{1-L}$ bottle of stock ${\text{HNO}}_{3}$ solution (we could assume any volume, the answer must always be the same - in this case, 1L makes the calculations easier).

Knowing that the solution's density is $\text{1.41 g/mL}$, which is equal to

$1.41 \text{g"/color(red)(cancel(color(black)("mL"))) * (1000 color(red)(cancel(color(black)("mL"))))/"1 L" = "1410 g/L}$

we can determine the stock solution's mass

$\text{density" = "mass"/"volume}$

so you get

$\text{mass" = "density" * "volume}$

${m}_{\text{sol" = 1410"g"/color(red)(cancel(color(black)("L"))) * 1color(red)(cancel(color(black)("L"))) = "1410 g}}$.

Out of this quantity, 70.3% represents $H N {O}_{3}$, so nitric acid's mass will be

1410 color(red)(cancel(color(black)("g solution"))) * ("70.3 g HNO"""_3)/(100color(red)(cancel(color(black)("g solution")))) = "991.2 g HNO"""_3

Knowing that nitric acid's molar mass is $\text{63 g/mol}$, we determine the number of moles to be

991.2 color(red)(cancel(color(black)("g"))) * "1 mole"/(63 color(red)(cancel(color(black)("g")))) = "15.7 moles"

Thus, its molarity is

$C = \frac{n}{V} = \text{15.7 moles"/"1 L" = "15.7 M}$

We can now apply dilution concentration calculations to determine what the requested volume will be

${C}_{1} {V}_{1} = {C}_{2} {V}_{2}$

V_1 = C_2/C_1 * V_2 = (0.5 color(red)(cancel(color(black)("M"))))/(15.7color(red)(cancel(color(black)("M")))) * 1.5 * 10^3"mL" = color(green)("47.8 mL")#.

I'll leave the answer rounded to three sig figs.