Question

The percentage purity of commercial `"HCl"` is 35% (w/w) & the specific gravity of the solution is 1.18 `"g"//"mL"`. What are the molarity, molality and normality?

Answer 1

I happen to know (from using the following) that `37%` reaction-grade `"HCl"` is about `"12.06 M"`... so let's go off of that. I also looked that up here. You can also go off of `36%` `"HCl"` which is about `"11.65 M"` (easily googlable).

It's dissolved in water, and `"HCl"` is

`35.453+1.0079 = "36.4609 g/mol"`.

`"1 L"` of the solution is

`"1.18 g"/cancel"mL"cdot (1000cancel("mL"))/cancel("1 L")cdotcancel"1 L" = "1180 g"`.

Thus, `35"% w/w"` for `"HCl"` puts it at `"413 g HCl"/"1180 g soln"`.

Molarity is `("mol solute")/("L solution")`, thus it is:

`"Molarity" = [413cancel("g")*("1 mol")/(36.4609cancel("g"))]/("1 L") ~~ color(blue)("11.3272 M")`

which is sensible considering how `37%` is close to `"12 M"`.

Molality is `("mol solute")/("kg solvent")`.

The density of water at `25^@"C"` is `"0.9970749 g/mL"`. Thus, with `"1 L"` of water, it has a mass of `"0.9970749 kg"`. Then:

`"Molality" = [413cancel("g")*("1 mol")/(36.4609cancel("g"))]/("0.9970749 kg") ~~ color(blue)("11.3604 m")`

Since `HCl` is monoprotic, its normality is equal to its molarity. Thus, its normality is also:

`"Normality"` `~~` `color(blue)("11.3272 N")`