At 25.0∘c, the molar solubility of silver chromate in water is 1.10×10−12 m . calculate the solubility in grams per liter. express your answer in grams per liter to three significant figures. how many liters of water are required to dissolve 1.00 g of silver chromate? express your answer in liters to three significant figures.

Solubility is and of water is required to dissolve 1 g of silver chromate.

Explanation: The equation for the reaction will be as follows:

1 mole of gives 2 moles of and 1 mole of

Thus if solubility of is s moles/liter, solubility of is 2s moles\liter and solubility of is s moles/liter

Given:-

Solubility in grams/liter=

Solubility in grams/liter=

of dissolve in= 1L

1 g of dissolve in=

The solubility silver chromate in grams per liter = 3.65. 10⁻¹⁰

2,74.10⁹ liters of water are required to dissolve 1.00 g of silver 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 of the AxBa solution can be stated as follows.

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

s                                xs              ys

Ksp = [Aᵃ⁺] ˣ [Bˣ⁻] ᵃ

Ksp = (xs) ˣ (ys) ᵃ

Solubility unit in the form of mole / liter or gram / liter

Solubility of Ag₂CrO₄ = 1.10 .10⁻¹² mole / liter

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

M Ag₂CrO₄ = 2.Ar Ag + Ar Cr + 4. Ar O

M Ag₂CrO₄ = 2.108 + 52 + 4.16

M Ag₂CrO₄ = 332

So the solubility is in grams / liter

= 1.10. 10⁻¹² x 332

= 3.65. 10⁻¹⁰

(3 significant numbers, 3.6 and 5)

If we dissolve 1 gram of silver chromate into the solution, we need as much water as possible

= 1 gram / 3.65. 10⁻¹ grams / liter

= 2.74.10⁹ liter

(3 significant numbers, 2,7 and 4)

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