The molar solubility is 7.4×
M and the solubility is 7.4× g/L .
Calculation ,
The dissociation of silver bromide is given as ,
→ 
+ 
S
- S S
Ksp = [ ] [ ] = [S] [ S ] = 
S = √ Ksp = √ 5. 5×
= 7.4×
The solubility =7.4× g/L
The molar solubility is the solubility of one mole of the substance.
Since , one mole of is dissociates and form one mole of each and ion . So, solubility is equal to molar solubility but unit is different.
Molar solubility = 7.4× mol/L = 7.4× M
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Explanation:
55.9 kPa; Variables given = volume (V), moles (n), temperature (T)
We must calculate <em>p</em> from <em>V, n</em>, and <em>T</em>, so we use <em>the Ideal Gas Law</em>:
<em>pV = nRT</em>
Solve for <em>p</em>: <em>p = nRT/V</em>
R = 8.314 kPa.L.K^(-1).mol^(-1)
<em>T</em> = (265 + 273.15) K = 538.15 K
<em>V</em> = 500.0 mL = 0.5000 L
∴ <em>p</em> = [6.25 x 10^(-3) mol x 8.314 kPa·L·K^(-1)·mol^(-1) x 538.15 K]/(0.5000 L) = 55.9 kPa
Answer:
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