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nignag [31]
1 year ago
8

Calculate the molar solubility of silver(i) bromate with ksp = 5. 5×10-5. also, convert the molar solubility to the solubility.

the molar solubility is ___________ and the solubility is ____________
Chemistry
1 answer:
Sindrei [870]1 year ago
7 0

The molar solubility is 7.4×10^{-3} M and the solubility is  7.4×10^{-3} g/L .

Calculation ,

The dissociation of silver bromide is given as ,

AgBr → Ag ^{+} + Br^{-}

S  

 -          S        S        

Ksp =  [Ag ^{+} ] [ Br^{-} ]  =  [S] [ S ] = S^{2}

S = √ Ksp = √ 5. 5×10^{-5} = 7.4×10^{-3}

The solubility =7.4×10^{-3} g/L

The molar solubility is the solubility of one mole of the substance.

Since ,  one mole of AgBr is dissociates and form one mole of each  Ag ^{+} and Br^{-} ion . So, solubility is equal to molar solubility but unit is different.

Molar solubility = 7.4×10^{-3} mol/L = 7.4×10^{-3} M

To learn more about molar solubility ,

brainly.com/question/16243859

#SPJ4

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Answer:

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Explanation:

1. Mass / Molar mass = Mol

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3. 1 mol of anything, has 6.02x10²³ particles

6.02x10²³ molecules . 1.2 mol = 7.22x10²³

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1 amu = 1.66054x10⁻²⁴g

5.24x10²⁵ amu = (5.24x10²⁵ . 1.66054x10⁻²⁴) = 89.6 g

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0.0222 moles → ( 0.0222 . 6.02x10²³) = 1.34x10²²

6. Density of water is 1 g/mL, so 500 mL are contained in 500 g of water

Molar mass H₂O = 18 g/m

500 g / 18 g/m = 27.8 moles

6.02x10²³ molecules . 27.8 moles = 1.67x10²⁵

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