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2 years ago

The sp of pbbr2 is 6. 60×10−6. what is the molar solubility of pbbr2 in pure water?

Chemistry

1 answer:

Inessa05 [86]2 years ago

6 0

Ksp of PbBr₂ is 6.60 × 10⁻⁶. The molar solubility of PbBr₂ in pure water is 0.0118M.

Ksp or Solubility Product Constant is an equilibrium constant for the dissociation in an aqueous solution.

Molar solubility (S) is the concentration of the dissolved substance in a solution that is saturated.

Let the molar solubility be S upon dissociation.

PbBr₂ or Lead Bromide dissociates in pure water as follows:

PbBr₂ ----------> Pb⁺² + Br⁻

S 2S

Ksp = [Pb⁺²] [ Br⁻]

Ksp = (S) (2S)²

Ksp = 4S³

6.60 × 10⁻⁶ = 4S³

S = 0.0118M

Hence, the Molar solubility S is 0.0118M.

Learn more about Molar solubility here, brainly.com/question/16243859

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An automobile antifreeze mixture is made by mixing equal volumes of ethylene glycol (d = 1.114 g/mL; M = 62.07 g/mol) and water

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(a) The volume percent is, 50.63 %

(b) The mass percent is, 52.69 %

(c) Molarity is, 9.087 mole/L

(d) Molality is, 17.947 mole/L

(e) Moles fraction of ethylene glycol is, 0.244

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Density of ethylene glycol = 1.114 g/mL

Molar mass of ethylene glycol = 62.07 g/mole

Density of water = 1.00 g/mL

Density of solution or mixture = 1.070 g/mL

According to the question, the mixture is made by mixing equal volumes of ethylene glycol and water.

Suppose the volume of each component in the mixture is, 1 mL

First we have to calculate the mass of ethylene glycol.

$\text{Mass of ethylene glycol}=\text{Density of ethylene glycol}\times \text{Volume of ethylene glycol}=1.114g/mL\times 1mL=1.114g$

Now we have to calculate the mass of water.

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1.00g/mL\times 1mL=1.00g$

Now we have to calculate the mass of solution.

Mass of solution = Mass of ethylene glycol + Mass of water

Mass of solution = 1.114 + 1.00 = 2.114 g

Now we have to calculate the volume of solution.

$\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=\frac{2.114g}{1.070g/mL}=1.975mL$

(a) Now we have to calculate the volume percent.

$\text{Volume percent}=\frac{\text{Volume of ethylene glycol}}{\text{Volume of solution}}\times 100=\frac{1mL}{1.975mL}\times 100=50.63\%$

(b) Now we have to calculate the mass percent.

$\text{Mass percent}=\frac{\text{Mass of ethylene glycol}}{\text{Mass of solution}}\times 100=\frac{1.114g}{2.114g}\times 100=52.69\%$

(c) Now we have to calculate the molarity.

$\text{Molarity}=\frac{\text{Mass of ethylene glycol}\times 1000}{\text{Molar mass of ethylene glycol}\times \text{Volume of solution (in mL)}}$

$\text{Molarity}=\frac{1.114g\times 1000}{62.07g/mole\times 1.975L}=9.087mole/L$

(d) Now we have to calculate the molality.

$\text{Molality}=\frac{\text{Mass of ethylene glycol}\times 1000}{\text{Molar mass of ethylene glycol}\times \text{Mass of water (in g)}}$

$\text{Molality}=\frac{1.114g\times 1000}{62.07g/mole\times 1kg}=17.947mole/kg$

(e) Now we have to calculate the mole fraction of ethylene glycol.

$\text{Mole fraction of ethylene glycol}=\frac{\text{Moles of ethylene glycol}}{\text{Moles of ethylene glycol}+\text{Moles of water}}$

$\text{Moles of ethylene glycol}=\frac{\text{Mass of ethylene glycol}}{\text{Molar of ethylene glycol}}=\frac{1.114g}{62.07g/mole}=0.01795mole$

$\text{Moles of water}=\frac{\text{Mass of water}}{\text{Molar of water}}=\frac{1g}{18g/mole}=0.0555mole$

$\text{Mole fraction of ethylene glycol}=\frac{0.01795mole}{0.01795mole+0.0555mole}=0.244$

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