The solubility of a salt of molar mass 101g at 20oC is 0.34 mol dm-3. If 3.40g of the salt is dissolved completely in 250cm3 of water in a beaker, the resulting solution is
A. A suspension B. Saturated C. Unsaturated D. Supersaturated
Correct Answer: C
Explanation
At 20oC 0.34 mol/dm of the salt dissolve = 0.34 * 101 = 34.34gm Of the salt dissolved. ∴ 34.34gm dissolve is 1,000cm3 If only 250 cm3 is the required mass X = (250/1000) * 34.34 = 8.555gm If only 3.40gm of the salt dissolved ,the solution is unsaturated
