Question

Question #e73b2

Answer 1

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"` `HNO_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"` `HNO_3`

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

`C = n/V = "15.73 moles"/"1.00 L" = "15.73 M"`

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

`C_1V_1 = C_2V_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 * 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 = 0.079cancel("L") * "1000 mL"/(1cancel("L")) = color(green)("79 mL")`

Answer 2

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 `HNO_3` solution.

`c=n/v`

So:

`n=cv`

So the no. moles `HNO_3` required = `2.5xx0.5=1.25`

Now we need to find how many grams this is. We get this by multiplying by the weight of 1 mole of `HNO_3` which is equal to its `M_r` in grams.

`M_r[HNO_3]=63.01`

So mass required = `1.25xx63.01=78.76"g"`

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

This means that 100g of this concentrated solution contains 70.30g of `HNO_3`.

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

There are:

`70.30"g"` `HNO_3` in `100"g"` of concentrated solution.

`1"g"` is present in `(100)/(70.30)"g"` of concentrated solution.

So:

`78.76"g"` is present in `(100)/(70.30)xx78.76=112.03"g"` of concentrated solution.

The density of the solution = 1.41g/ml

density = mass/volume

So volume = mass/density

So the volume required = `(112.03)/(1.41)=79.45"ml"`