Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.

# An aqueous solution of ##"HCl"## contains 36% ##"HCl"## by mass. The density of the solution is 1.189 g/ml. What is the mole fraction of ##"HCl"## in the solution ? What is the molality of ##"HCl"## in the solution ? Show all calculations

Here's what I got.

Interestingly enough, you don't need to know the of the solution to calculate its and the **mole fraction** of hydrochloric acid, ##"HCl"##, your .

All you need to know here is the solution's mass by mass , ##"% m/m"##, which is said to be equal to ##36%##.

Now, to make the calculations easier, pick a sample of this solution that has a mass of ##"100 g"##. According to the given percent concentration, this sample will contain

- ##"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, ##chi_"HCl"##, is defined as the number of moles of acid divided by the **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 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 , 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)|)))##