# Question 00785

Oct 12, 2015

For part (b) $V = \text{0.898 L}$

#### Explanation:

I'll show you how to solve part (b), since it's a little more interesting than part (a).

So, you know that you're dealing with a solution of potassium dihydrogen arsenate, ${\text{KH"_2"AsO}}_{4}$, of known molarity.

You also know that molairty is defined as moles of solute, which in you case is potassium dihydrogen arsenate, divided by liters of solution.

$\text{molarity" = "moles of solute"/"liters of solution"" }$, or $\text{ "C = n_"solute"/V_"solution}$

So, you need to pick out a volume of this $\text{0.0500 M}$ solution, but you don't know how many moles of solute it must contain.

Notice that they give you the mass of the solute instead. This means that you're going to have to calculate how many moles of potassium dihydrogen arsenate you ahve in that many grams of the compound.

To do that, use its molar mass, which tells you what the mass of one mole of a substance is

8.10color(red)(cancel(color(black)("g"))) * ("1 mole KH"_2"AsO"_4)/(180.0334color(red)(cancel(color(black)("g")))) = "0.04492 moles"

Now you know how many moles of solute you need to get from the sample. Rearrange the equation to solve for the volume that would contain this many moles

$C = \frac{n}{V} \implies V = \frac{n}{C}$

V = (0.04492color(red)(cancel(color(black)("moles"))))/(0.0500color(red)(cancel(color(black)("moles")))/"L") = "0.8984 L"#

You need to round this off to three sig figs, the number of sig figs you have for the molarity of the solution and mass of the solute

$V = \textcolor{g r e e n}{\text{0.898 L}}$