Chemistry, 22.01.2020 04:31 carolinerosewillis

Asolution is prepared by dissolving 38.6 g sucrose (c12h22o11) in 495 g of water. determine the mole fraction of sucrose if the final volume of the solution is 508 ml.

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The mole fraction of sucrose in the final solution prepared by dissolving 38.6 g sucrose ( C₁₂H₂₂O₁₁) in 495 g of water is 0.004

We'll begin by calculating the number of mole sucrose and water in the solution. This can be obtained as follow:

For sucrose:

Molar mass of C₁₂H₂₂O₁₁ = (12×12) + (1×22) + (16×11)

= 144 + 22 + 176

= 342 g/mol

Mass of C₁₂H₂₂O₁₁ = 38.6 g

Mole of C₁₂H₂₂O₁₁ =?

Mole = mass / molar mass

Mole of C₁₂H₂₂O₁₁ = 38.6/342

Mole of C₁₂H₂₂O₁₁ = 0.113 moleFor water:

Molar mass of H₂O = (2×1) + 16

= 18 g/mol

Mass of H₂O = 495 g

Mole of H₂O =?

Mole = mass / molar mass

Mole of H₂O = 495 / 18

Mole of H₂O = 27.5 mole

Next, we shall determine the total mole.

Mole of C₁₂H₂₂O₁₁ = 0.113 mole

Mole of H₂O = 27.5 mole

Total mole =?

Total mole = 0.113 + 27.5

Total mole = 27.613 mole

Finally, we shall determine the mole fraction of sucrose.

Mole of Sucrose, C₁₂H₂₂O₁₁ = 0.113 mole

Total mole = 27.613 mole

Mole fraction of sucrose =?

Mole fraction = mole of element / total mole

Mole fraction of sucrose = 0.113 / 27.613

Mole fraction of sucrose = 0.004

Therefore, the mole fraction of sucrose in the solution is 0.004

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Answer from: ethangeibel93

Mole fraction is defined as the ratio of number of moles of a constituent of a solution to the total amount of all constituent in a mixture. The mole fraction of sucrose in 508 ml of solution will be 0.004.

Given that,

Molar mass of C $_{12}$ H $_{22}$ O $_{11}$ = 144 + 22 + 176 = 342 g/molGiven Mass of C $_{12}$ H $_{22}$ O $_{11}$ = 38.6 g $\text{moles}&=\dfrac{\text{given mass}}{\text{molar mass}}$ $\text{moles}&=\dfrac{38.6}{342}$

Moles = 0.113

Similarly, for water:

Molar Mass of Water = 2 + 16 = 18 g/molGiven Mass of Water = 495 g $\text{moles}&=\dfrac{\text{given mass}}{\text{molar mass}}$ $\text{moles}&=\dfrac{495}{18}$

Moles = 27.5

Now, total moles of the solution is:

Total moles = 0.113 + 27.5

Total moles = 27.613 mole

Calculating the mole fraction, we get:

$\text{Mole Fraction}&=\dfrac{\text{Moles of sucrose}}{\text{Total moles}}\\\\\text{Mole Fraction}&=\dfrac{0.113}{27.613}$ Mole fraction of sucrose = 0.004

Therefore, the mole fraction of sucrose for the given solution is 0.004.

To know more about mole fraction, refer to the following link:

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Answer from: Quest

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction: so₂cl₂ (g) = so₂ (g) + cl₂ (g) total

the moles at equilibrium: 1 - α α α = 1+α

concentrations: 1 - α α α = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

$k_{c}$ = $\frac{[so2cl2] x [cl2]}{[so2cl2]}$

or, $\frac{\alpha ^{2} }{(1-\alpha)v }$

or, $k_{c}$ = $\frac{\alpha ^{2} }{(1-\alpha) }$ (at v = 1l)

the reaction: so₂cl₂ (g) = so₂ (g) + cl₂ (g) total

the moles at equilibrium: 1 - α α α = 1+α

concentrations: (1 - α)/0.5 α/0.5 α/0.5 = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

$\frac{[so2cl2] x [cl2]}{[so2cl2]}$

or, $\frac{\alpha ^{2} }{(1-\alpha)0.5 }$

$\frac{\alpha ^{2} }{(1-\alpha) }$ = $\frac{\alpha ^{2} }{(1-\alpha) 0.5}$

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%

Answer from: Quest

i think the answer is c hope i'm right

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