Question

A chemistry student is given 3.00 L of a clear aqueous solution at 17 C . He is told an unknown amount of a certain compound X is dissolved in the solution. The student allows the solution to cool to 17 C . The solution remains clear. He then evaporates all of the water under vacuum. A precipitate remains. The student washes, dries and weighs the precipitate. It weighs 0.15 kg . Using the above information can you calculate the solubility, X, in water at 17 C.

Answer 1

The right approach is "50 g/l".

Step-by-step explanation:

The given values are:

Mass or solute or precipitation,

= 0.15 kg

on converting it into "g", we get

=
`0.15 \ kg* (1000 \ g)/(1 \ kg)`

=
`150 \ g`

Volume of solution,

= 3.00 L

Now,

The solubility of X will be:

=
`(Mass \ of \ X)/(Volume \ of \ solution)`

On substituting the values, we get

=
`(150)/(3)`

=
`50 \ g /l`