How many liters of water are required to dissolve 1.00 g of barium chromate? express your answer in liters to three significant figures?

The solubility  Barium chromate in grams per liter = 2.78 . 10⁻³ grams/L

359 liters of water are required to dissolve 1.00 g of Barium chromate

Solubility is the maximum amount of a substance that can dissolve in some solvents. Factors that affect solubility

Ksp is an ion product in equilibrium

Solubility relationships and solubility constants (Ksp) of the AxBa solution can be stated as follows.

AₓBₐ (s) ← ⎯⎯⎯⎯ → x Aᵃ⁺ (aq) + a Bˣ⁻ (aq)

s                             xs               as

Ksp = [Aᵃ⁺] ˣ [Bˣ⁻] ᵃ

Ksp = (xs) ˣ (as) ᵃ

Solubility units in the form of mol / liter or gram / liter

At 25.°C, the molar solubility of Barium chromate  BaCrO₄ in water is 1.10. 10⁻⁵M.

to change units to grams / liter, we multiply by molar mass:

M BaCrO₄ = Ba + Cr + 4. Ar O

M BaCrO₄ = 137 + 52 + 4.16

M BaCrO₄ = 253

So the solubility is in grams / liter

= 1.10 . 10⁻⁵ mol / liter x 253 grams / mol

=  278.3 .10⁻⁵ = 2.78 . 10⁻³ grams/L

(3 significant numbers, 2.7 and 8)

If we dissolve 1 gram of Barium chromate into the solution, we need water :

= 1 grams / 2.78 . 10⁻³ grams / liter

= 359 liters

(3 significant numbers, 3.5 and 9)

the rate of solubility

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influencing factors are both solubility and the rate of dissolution

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The solubility of a substance

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Keywords: solubility, silver chromate, a significant number

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

the dependant variable is spilled water

the independent is the heat