Question

A saturated solution of KClO3 is dissolved in 100 mL of water. If the saturated solution is cooled from 90oC to 60oC, how many grams of precipitate will be formed?

Answer 1

24 g

Step-by-step explanation:

To find the mass of precipitate that will be formed we need to use the solubility curve of KClO₃ in the water. From that graphic of solubility curve of KClO₃ in water, we have that:

The mass of the saturated solution of KClO₃ in 100 mL of water at 90 °C is = 50 g

The mass of the saturated solution of KClO₃ in 100 mL of water at 60 °C is = 26 g

Hence, by the difference of the mass of KClO₃ between 90 °C and 60 °C we can find the grams of precipitate that will be formed:

`m_(f) = m_((90^(\circ) C)) - m_((60 ^(\circ) C)) = 50 g - 26 g = 24 g`

Therefore, 24 g of KClO₃ precipitate will be formed.

I hope it helps you!