Question

Question #1c1fe

Answer 1

`12.15%`

Explanation:

The idea here is that you need to use the molar volume of a gas at STP to determine how many moles of hydrogen chloride, `"HCl"`, you're adding to the known mass of water.

It's worth mentioning that hydrochloric acid is the term used for aqueous solutions of hydrogen chloride. Simply put, `"HCl"` is known as hydrogen chloride in the gaseous state and as hydrochloric acid in aqueous solution.

`overbrace("HCl"_ ((g)))^(color(blue)("hydrogen chloride")) -> overbrace("HCl"_ ((aq)))^(color(red)("hydrochloric acid"))`

Once you know how many moles of hydrochloric acid are present in solution, use the compound's molar mass to convert it to grams.

So, the old definition of STP conditions imply a pressure of `"1 atm"` and a temperature of `0^@"C"`. Under these conditions, one mole of any ideal gas occupies `"22.4 L"`.

This means that your sample, which is kept at STP conditions, will contain

`22.4 color(red)(cancel(color(black)("L"))) * overbrace(("1 mole HCl"_ ((g)))/(22.4color(red)(cancel(color(black)("L")))))^(color(purple)("molar volume of a gas at STP")) = "1 mole HCl" _((g))`

As you know, the molar mass tells you the mass of one mole of a given compound. In this case, hydrogen chloride's molar mass of `"36.46 g mol"^(-1)` tells you that one mole of hydrogen chloride has a mass of `"36.46 g"`.

Since you're dealing with a `"1-mole"` sample, you will end up with `"36.46 g"` of hydrochloric acid in solution.

The total mass of the sample will now be

`m_"total" = m_"water" + m_"HCl"`

`m_"total" = "263.5 g" + "36.46 g" = "299.96 g"`

This means that the percent concentration by mass of hydrochloric acid will be

`"% HCl" = (36.46 color(red)(cancel(color(black)("g"))))/(299.96color(red)(cancel(color(black)("g")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)12.15%color(white)(a/a)|)))`

I'll leave the answer rounded to four sig figs.