# Question #7d76d

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

Interestingly enough, you don't need to know the density of the solution to calculate its **molality** and the **mole fraction** of hydrochloric acid,

All you need to know here is the solution's mass by mass percent concentration,

Now, to make the calculations easier, pick a sample of this solution that has a *mass* of

#"36 g"# of hydrochloric acid#"64 g"# of water

Use the **molar masses** of hydrochloric acid and of water to determine how many *moles* of each you have in this sample

#36 color(red)(cancel(color(black)("g"))) * "1 mole HCl"/(36.46color(red)(cancel(color(black)("g")))) = "0.9874 moles HCl"#

#64 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "3.553 moles H"_2"O"#

The **mole fraction** of hydrochloric acid, **total number of moles** present in solution.

In your case, you will have

#chi_"HCl" = (0.9874 color(red)(cancel(color(black)("moles"))))/((0.9874 + 3.553)color(red)(cancel(color(black)("moles")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(0.22)color(white)(a/a)|)))#

Next, the **molality** of th solution is defined as the number of moles of solute divided by the **mass of solvent expressed in kilograms**. In your case, the molality of the solution will be

#b = "0.9874 moles"/(64 * 10^(-3)"kg") = "15.4 mol kg"^(-1)#

I'll round this off to two **sig figs**, the number of sig figs you have for the solution's percent concentration.

#b = color(green)(|bar(ul(color(white)(a/a)color(black)("15 mol kg"^(-1))color(white)(a/a)|)))#