The solubility of a solute in water is dependent on temperature as given by- S-A e - H-RT- where - H- heat of solution-Solute - H 2 Ol- Solution- - H- xFor a given solution- the variation of logS with temperature shown graphically- Hence- the solute is-

The correct option is D CaO S = Ae^ Δ H/RT log S = Δ HRT + log A slope = Δ HR gt; 0 Δ H lt; 0 Therefore the r × n CaO ...

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Standard VIII

Chemistry

Ways to Define Concentration

The solubilit...

Question

The solubility of a solute in water is dependent on temperature as given by:
$S = A e^{- Δ H / R T}$ , where $Δ H =$ heat of solution.
Solute $+ H_{2} O (l) ⇌$ Solution; $Δ H = \pm x$
For a given solution, the variation of $log S$ with temperature shown graphically. Hence, the solute is:

C a {S O}_{4} \cdot 5 H_{2} O

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N a C l

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sucrose

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C a O

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Solution

The correct option is D $C a O$
$S = A e^{- Δ H / R T}$
$log S = \frac{- Δ H}{R T} + log A$
slope $= \frac{- Δ H}{R} > 0$
$Δ H < 0$
Therefore the $r \times n$
$C a O + H_{2} O ⇌ C a (O H)_{2}; Δ H < 0$
is the only $r \times n$ among the given solute which is exothermic $(Δ H < 0)$ in nature.

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Similar questions

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}

, where

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 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 ?

H_{2 (g)} + \frac{1}{2} O_{2 (g)} \to H_{2} O_{(l)}

;

Δ H = - 286.2 k J

;

H_{2} O_{(l)} \to H_{(a q)}^{+} + O H_{(a q)}^{-}

;

Δ H = + 57.3 k J

. Enthalpy of ionisation of

O H^{-}

in aqueous solution is:

Q. Using the following data, mark the option(s) where

Δ H

is correctly written for the given reaction.

Given that :

H^{+} (a q) + {O H}^{-} (a q) ⟶ H_{2} O (l); Δ H = - 57.3 k J

{Δ H}_{s o l u t i o n}

H A (g) = - 70.7 k J / m o l

{Δ H}_{s o l u t i o n}

B O H (g) = 20 k J / m o l

{Δ H}_{i o n i z a t i o n}

H A = 15 k J / m o l

and

B O H

is a strong base.

At a particular temperature $H^{+} (a q) + O H^{-} (a q) \to H_{2} O (l); Δ H = - 57.1 k J$ . The approximate heat evolved when $400 m L$ of $0.2$ $H_{2} S O_{4}$ is mixed with $600 m L$ of $0.1 m$ $K O H$ solution will be:

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The solubility of a solute in water is dependent on temperature as given by:S=Ae−ΔH/RT, where ΔH= heat of solution.Solute +H2O(l)⇌ Solution; ΔH=±xFor a given solution, the variation of logS with temperature shown graphically. Hence, the solute is:

The correct option is D CaOS=Ae−ΔH/RTlogS=−ΔHRT+logAslope =−ΔHR>0ΔH<0Therefore the r×nCaO+H2O⇌Ca(OH)2;ΔH<0is the only r×n among the given solute which is exothermic (ΔH<0) in nature.

The solubility of a solute in water is dependent on temperature as given by:
$S = A e^{- Δ H / R T}$ , where $Δ H =$ heat of solution.
Solute $+ H_{2} O (l) ⇌$ Solution; $Δ H = \pm x$
For a given solution, the variation of $log S$ with temperature shown graphically. Hence, the solute is: