JAMB - Chemistry (2002 - No. 14)
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 suspension
Saturated
Unsaturated
Supersaturated
Explication
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
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