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You have prepared a saturated solution of x at 20∘c using 37.0 g of water. how much more solute can be dissolved if the temperature is increased to 30∘c?

Respuesta :

Erythrosin17
The solubility of the solute should be given here which should be as follows:

temperature at 20 degrees C : solubility 11.0 g X / 100 g H2O 
temperature at 30 degrees C : solubility 23.0 g X / 100 g H2O

To determine how much more of the solute can be added to the solution when the temperature is increased, we calculate the total mass of the solute that can be dissolved in the solution at 30 degrees Celsius.

Mass of X = 23.0 g X / 100 g H2O (37.0 g H2O) = 8.51 g X at T=30 degrees Celsius

Then, we calculate also the mass of X at 20 degrees Celsius

Mass of X = 11.0 g X / 100 g H2O ( 37.0 g H2O) = 4.07 g X

Mass of X that can be added after increasing the temperature = 8.51 - 4.07 = 4.44 g of X

To the prepared solution at  [tex]\rm 20^\circ C[/tex] the extra mass of x to be added to the solution at  [tex]\rm 30^\circ C[/tex] will be 4.44 g.

The complete question has given:

Solubility of x/100 g water at [tex]\rm 20^\circ C[/tex] = 11

Solubility of x/100 g water at [tex]\rm 30^\circ C[/tex] = 23

Now, since the saturated solution at [tex]\rm 20^\circ C[/tex]  will use 37 g of water. The mass of x in the saturated solution will be :

Mass of x =  [tex]\rm \dfrac{solubility\;of\;x\;\times\;mass\;of\;water}{100}[/tex]

Mass of x at [tex]\rm 20^\circ C[/tex] = [tex]\rm \dfrac{11\;\times\;37}{100}[/tex] g

Mass of x at [tex]\rm 20^\circ C[/tex]  = 4.07 g.

The mass of solute at [tex]\rm 30^\circ C[/tex] will be = [tex]\rm \dfrac{23\;\times\;37}{100}[/tex] g.

The mass of solute at[tex]\rm 30^\circ C[/tex]  = 8.51 g.

The extra mass of x to be added to the solution will be = 8.51 - 4.07 g.

The extra mass of x to be added to the solution will be 4.44 g.

For more information, refer to the link:

https://brainly.com/question/8678339

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