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
Explanation
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an easy method is using the formula
Solubility=mass/molarmass x1000/volume
solubility=0.34,molarmass=101,volume=250
then we find mass
0.34=mass/101 x 1000/250
mass=8.585g
then we can see than the mass that was dissolved is less than 8.585g which means it is unsaturated(c).

Salt X: 101 g/mol with a solubility of 0.34 mol/dm³.
Goes to show that the mass of salt X that will completely saturate 1 dm³ of solvent, with respect to its molar mass and solubility value, is given by:
saturatory mass = solubility value x molar mass
: 0.34 x 101 = 34.34g.
{} if 34.34g can completely saturate 1000 cm³,
then () can completely saturate 250 cm³.
:. () = [250 cm³ x 34.34 g] ÷ 1000 cm³
=> 8.585 g only, is needed to completely saturate 250 cm³ of solvent.
The question indicated that only 3.4g was dissolved in the 250 cm³ solvent; which is less than saturatory mass, so the solution is still unsaturated. (that guhhdamn temperature is a redundant and useless information, UTME will not confuse you in Jesus Name
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