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: Option 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
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