Question

A 2.0 L kettle contains 116 g of boiler scale (CaCO3). How many times would the kettle have to be completely filled with distilled water
to remove all of the deposit at 25 ec, of CaCO3 = 87x

Answer 1

Answer:

big answer took 20 mins to write it .mark as brainliest for my hard work :)

Explanation:

From Ksp you can calculate the solubility of CaCO3.

In dissociation of one CaCO3 molecule one calcium ion and one carbonate ion are formed:  

CaCO₃←→ Ca²⁺ + CO₃²⁻  

so the ion concentrations are equal to the apparent concentration of dissolved calcium carbonate

[Ca²⁺] = [CO₃²⁻] = [CaCO₃]

At maximum solubility ion concentrations satisfy solubility product:

Ksp = [Ca²⁺] · [CO₃²⁻]

Hence

Ksp = [CaCO₃]²

<=>

[CaCO₃] = √Ksp

= √8.7×10-9

= 9.237×10-5 M  

Multiply by molar mass to get the maximum soluble mass per unit volume:

ρ(CaCO₃) = [CaCO₃] · M(CaCO₃)

=  9.237×10-5mol/L · 100.087 g/mol

= 9.335×10-3g/L

So the minimum amount of water needed to dissolve 116g is

V  = 116g / 9.335×10-3g/L = 12,426L