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What is 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?

Sagot :

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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