# Question 2e118

Nov 6, 2015

225 mL

#### Explanation:

This is a basic (M1)(V1)=(M2)(V2) problem. The only variable that you're missing is the M2.
To find that you have to solve the molarity problem. When given a percentage you can assume that you have that many grams of solute in 100 grams of solvent.

Therefore:
37g HCl X 1 mol HCl X 1.20 g soln X 1000mL
100 g soln 36.458 g HCl 1 ml soln 1 L

This will tell you the molarity of the stock solution, it's important to see that 1.20 g/mL is your conversion factor.

Once you solve the dimensional analysis you can plug that in as your M2 variable.

it will look like (.600)(4.5)=(12)(V2)
Some simple algebra will give you the answer in L. Multiply that by 1000 to get it in mL.

Nov 7, 2015

$\text{220 mL}$

#### Explanation:

The idea here is that you need to pick a sample of the stock solution, find its molarity, then use the appropriate number of moles of hydrochloric acid to make the target solution.

So, to make calculations easier, let's say that we want to pick a $\text{1.00-L}$ sample of the stock hydrochloric acid solution.

Use the given density to determine what the mass of this sample will be

1.00color(red)(cancel(color(black)("L"))) * (1000color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "1.20 g"/(1color(red)(cancel(color(black)("mL")))) = "1200 g"

You know that the stock solution has a percent concetration by mass of hydrochloric acid equal to 37%. This means that every $\text{100 g}$ of solution will contain $\text{37 g}$ of hydrochloric acid.

In your case, this sample will contain

1200color(red)(cancel(color(black)("g solution"))) * "37 g HCl"/(100color(red)(cancel(color(black)("g solution")))) = "444 g HCl"

Now use hydrochloric acid's molar mass to find the number of moles you have in that many grams

444color(red)(cancel(color(black)("g"))) * "1 mole HCl"/(36.46color(red)(cancel(color(black)("g")))) = "12.178 moles HCl"

SInce the sample has a volume of $\text{1.00 L}$, it follows that its molarity will be equal to

$c = \frac{n}{V} = \text{12.178 moles"/"1.00 L" = "12.2 M}$

Now, let's say that you don't know the equation for dilution calculations to determine what volume of stock solution would be neede to make the target solution.

Use the molarity and volume of the target solution to determine how many moles of hydrochloric acid it must contain

$c = \frac{n}{V} \implies \frac{n}{c} \cdot V$

$n = \text{0.600 M" * "4.5 L" = "2.70 moles HCl}$

Now, if $\text{1.00 L}$ of the stock solution contain $12.178$ moles of hydrochloric cid, waht volume would contain $2.70$ moles?

2.70color(red)(cancel(color(black)("moles HCl"))) * "1.00 L stock"/(12.178color(red)(cancel(color(black)("moles HCl")))) = "0.2217 L"

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

V_"stock" = 0.2217color(red)(cancel(color(black)("L"))) * (10^3"mL")/(1color(red)(cancel(color(black)("L")))) = "221.7 mL" = color(green)("220 mL")#

SIDE NOTE I recommend using different values for the initial volume of the sample just to make sure that the result must always come out to be 220 mL.