# Question e73b2

Jun 15, 2015

You need 79 mL of the stock solution.

#### Explanation:

The first thing you need to do is figure out exactly how much nitric acid your stock solution contains. To make the calculations easier, you can assume the you have a 1.00-L sample of the stock solution.

Use the solution's density to determine what the mass of the sample is

1.00cancel("L") * (1000cancel("mL"))/(1cancel("L")) * "1.41 g"/(1cancel("mL")) = "1410 g"

You know that this solution is 70.3% concentration by mass nitric acidm which means that every 100 g of solution contain 70.3 g of nitric acid. This means that you have

1410cancel("g") * ("70.3 g"HNO_3)/(100cancel("g")) = "991.23 g" $H N {O}_{3}$

Use nitric acid's molar mass to see how many moles would that many grams contain

991.23cancel("g") * ("1 mole"HNO_3)/(63.01cancel("g")) = "15.73 moles" $H N {O}_{3}$

This means that the 1.00-L sample will have a molarity of

$C = \frac{n}{V} = \text{15.73 moles"/"1.00 L" = "15.73 M}$

Now all you have to do is perform a simple dilution calculation

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

${C}_{1}$, ${V}_{1}$ - the molarity and volume of the stock solution sample;
${C}_{2}$, ${V}_{2}$ - the molarity and volume of the target solution.

Plug your values into the above equation and solve for ${V}_{1}$

${V}_{1} = {C}_{2} / {C}_{1} \cdot {V}_{2}$

V_1 = (0.500cancel("M"))/(15.73cancel("M")) * "2.5 L" = "0.07947 L"

Rounded to two sig figs, the number of sig figs you gave for the volume of the target solution, and expressed in mL, the answer will be

${V}_{1} = 0.079 \cancel{\text{L") * "1000 mL"/(1cancel("L")) = color(green)("79 mL}}$

Jun 15, 2015

You should use 79.45 ml of the stock solution and make it up to a volume of 2.5L

#### Explanation:

You need 2.5L of 0.5M $H N {O}_{3}$ solution.

$c = \frac{n}{v}$

So:

$n = c v$

So the no. moles $H N {O}_{3}$ required = $2.5 \times 0.5 = 1.25$

Now we need to find how many grams this is. We get this by multiplying by the weight of 1 mole of $H N {O}_{3}$ which is equal to its ${M}_{r}$ in grams.

${M}_{r} \left[H N {O}_{3}\right] = 63.01$

So mass required = $1.25 \times 63.01 = 78.76 \text{g}$

We know that the concentrated form is 70.30%# by mass.

This means that 100g of this concentrated solution contains 70.30g of $H N {O}_{3}$.

So now we need to work out what volume of concentrated solution will give us the mass we need:

There are:

$70.30 \text{g}$ $H N {O}_{3}$ in $100 \text{g}$ of concentrated solution.

$1 \text{g}$ is present in $\frac{100}{70.30} \text{g}$ of concentrated solution.

So:

$78.76 \text{g}$ is present in $\frac{100}{70.30} \times 78.76 = 112.03 \text{g}$ of concentrated solution.

The density of the solution = 1.41g/ml

density = mass/volume

So volume = mass/density

So the volume required = $\frac{112.03}{1.41} = 79.45 \text{ml}$