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?

1 Answer
Ernest Z.
Jul 6, 2016

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"

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