Question

What is the molality of a solution of phosphoric acid, `H_3PO_4` that contains 24.5 g of phosphoric acid (molar mass 98.0 g) in 100 g of `H_2O`?

Answer 1

The molality is 2.50 mol/kg.

Explanation:

The formula for molality `b` is

`color(blue)(|bar(ul(color(white)(a/a) b = "moles of solute"/"kilograms of solvent"color(white)(a/a)|)))" "`

`"Moles of H"_3"PO"_4 = 24.5 color(red)(cancel(color(black)("g H"_3"PO"_4))) × ("1 mol H"_3"PO"_4)/(98.0 color(red)(cancel(color(black)("g H"_3"PO"_4)))) = "0.250 mol H"_3"PO"_4`

`"Mass of solvent" = "100 g" = "0.100 kg"`

∴ `b = "0.250 mol"/"0.100 kg" = "2.50 mol/kg"`