The molality of phosphoric acid, h3po4, in a solution prepared from 376 g of phosphoric acid and 1. 70×103 g of ethylene glycol, c2h4(oh)2 is 2.25m.
Ethylene Glycol is known as C2H4 (OH)2.
It is used to prepare Antifreeze solution by adding water in it.
Given,
Mass of phosphoric acid = 376g
Mass of ethylene acid = 1.7 × 10 3g
Since ethylene glycol is in excess. So it is acts as a solvent and phosphoric acid acts as solute
As we know that,
Molar mass of phosphoric acid = 98g
Moles of phosphoric acid
= Given mass / Molar mass
=376 / 98
= 3.83
Molality is defined as the ratio of moles of solute to the mass of solvent.
Molality = Moles of solute / Mass of solvent
= 3.83 / 1.7 kg
= 2.25m.
Thus, we found that the molality of phosphoric acid is 2.25m.
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