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QUESTION

# 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)|)))