Question

what is the molality of ions in a .275 M solution of NH4)3PO4​

Answer 1

molarity= 0.275 M L-1

molarity= no. of mols of solute/ litres of solution

molality of NH4 + ions is required.

molality=no. of mols of NH4 ions/kilograms of solvent

Assuming, 1 mol of (NH4)3PO4 is dissolved.

mass of Ammonium present=3*(14+ 4*1) =>3*18 => 54 g.

So, 1 mol of (NH4)3PO4 has 54 grams of NH4+ in it or 3 mol of NH4+ .

Using formula for molarity, assuming 1 mol of (NH4)3PO4 is used.

0.275= 1/litres of solution

litres of solution= 1/0.275 => 40/11 litres.

40/11 litres of water dissolve 1 mol of Ammonium Phosphate.

40,000/11 cm3 of water is needed.

density of water is 1 g cm-3 (d=m/v)

mass= 40,000/11 g of water OR 40/11 kilograms of water.

So molality= 3 mol/ (40/11 kg of solvent)

=>33/40 mol kg-1

molality of NH4+ ions = 0.825 mol per kg of solvent.