Question

What is the solubility of 80 grams of magnesium chloride (MgCl2) dissolved into 150 cm3 of water?

Answer 1

S (MgCl2) = 0.533 g/mL

Step-by-step explanation:

Solubility ( S ) ≡ g/mL

⇒ m (MgCl2) = 80g

⇒ S (MgCl2) = 80g / 150 mL

⇒ S (MgCl2) = 0.533 g/mL

Answer 2

Solubility of magnesium chloride =
`533.33\;g/dm^3`

Step-by-step explanation:

Mass of
`MgCl_2` = 80 g

Volume of water =
`150\;cm^3`

Solubility is expressed in
`g/dm^3`

`1\;cm^3 = 0.001\;dm^3`

So,
`150\;cm^3 = 150* 0.001 = 0.15\;dm^3`

`Solubility = (Mass\;in\;g)/(Volume\;in\;dm^3)`

Substitute the values in the above formula,

`Solubility = (80\;g)/(0.15\;dm^3)=533.3\;g/dm^3`