1.

How many grams of potassium chloride is present in 250 g of saturated solution ? The solubility of KCl is 35.8 at 25^(@)C ?

Answer»

Solution :Solubility` = ("Mass of solute")/("Mass of solvent") XX 100`
`35.8 = (X)/(250 -x) xx 100`
`(250 - x)35.8 = 100 x rArr 8950 - 35.8 x = 100x rArr x = 65.9 G`


Discussion

No Comment Found