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Introduction to Le Chatelier's Principle

The solubilit...

Question

The solubility of a solute in water varies with temperature as given by $: S = A e^{\frac{- Δ H}{R T}}, Δ H$ being the enthalpy of solution. For a given solute, variation of In S with temperature is as shown in the figure. The solute is expected to be ?

A

$C a O$

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B

$C u S O_{4}$

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C

$M g S O_{4}$

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D

$C u S O_{4} .5 H_{2} O$

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Solution

The correct option is D
$C u S O_{4} .5 H_{2} O$

ln $S = l n A - \frac{Δ H}{R} \times \frac{1}{T}$
The slope of the graph is negative. So $Δ H > 0.$ Hence, dissolution is endothermic. Hydrated salt dissolves endothermically.

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The solubility of a solute in water varies with temperature as given by $: S = A e^{\frac{- Δ H}{R T}}, Δ H$ being the enthalpy of solution. For a given solute, variation of In S with temperature is as shown in the figure. The solute is expected to be ?

Q. The solubility of a solute in water is dependent on temperature as given by:

S = A e^{- Δ H / R T}

Δ H =

heat of solution.
Solute

+ H_{2} O (l) ⇌

Δ H = \pm x

For a given solution, the variation of

log S

with temperature shown graphically. Hence, the solute is:

Q. The solubility of a solute in a solvent (say

H_{2} O

) is dependent on temperature as given by

S = A e^{- Δ H / R T}

Δ H

is the heat of reaction. For a given solution, the variation of

l o g S

with temperature is shown graphically. Hence, solute is:

The amount of a solute present in its saturated solution at a given temperature is called the solubility of the solute at that temperature.

The solubility of an unknown substance in water at 50°C is 30g.

The solubility of the given substance in water at 100°C would be

Q. A saturated aqueous solution, is heated to a higher temperature. With increase in temperature, the solubility increases, but no additional solute is added to the solution. This condition is called as :

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