Question

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

Answer 1

Assuming the water is present at 20 °C, the solubility of barium chromate is 0.2275 mg/100 mL

Converting this to mg/L, we get:

2.275 mg/L

Converting this value again, to grams per liter, we get:

0.002275 g/L

To dissolve one gram, we require:

1/0.002275 = 439.56044

440 liters of water.

Answer 2

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

Further explanation

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

• 1. Temperature:

• 2. Surface area:

• 3. Solvent type:

• 4. Stirring process:

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)

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Keywords: solubility, silver chromate, a significant number