Question

salt water freezes at -2° C. how many grams of lithium phosphate would you need to add to a liter of water, to make it have the same boiling point as sea water

Answer 1

The elevation in boiling point upon adding of salt in water is given by

?T = change in boiling point
i = the van't Hoff factor (the number of particles into which the added solute dissociates)
k_b = the boiling point elevation constant for the solvent
m = molality

In case of Li₃PO₄, i= 8

For water k_b = 0.512 ^oC/molal

The average boiling point of sea water =104 ^oC

?T= 4

Now weight of 1 litre of water = 1kg.

Let the no. of moles of Lithium Phosphate required to increase the boilinig point be x.

= 0.98 moles

moleular weight of LiP= 115.79 gm

Weight of 0.98 moles = 113.47 gm

113.47 gm of Li₃PO₄, has to be added to 1 litre of water to make it have the same boiling point as sea water