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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.

Respuesta :

Cricetus

Answer:

The right approach is "50 g/l".

Explanation:

The given values are:

Mass or solute or precipitation,

= 0.15 kg

on converting it into "g", we get

= [tex]0.15 \ kg\times \frac{1000 \ g}{1 \ kg}[/tex]

= [tex]150 \ g[/tex]

Volume of solution,

=  3.00 L

Now,

The solubility of X will be:

=  [tex]\frac{Mass \ of \ X}{Volume \ of \ solution}[/tex]

On substituting the values, we get

=  [tex]\frac{150}{3}[/tex]

=  [tex]50 \ g /l[/tex]