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The solubility curve of $KN{O_3}$ in water is shown below. The amount of $KN{O_3}$ that dissolves in $50g$ of water at $40^\circ C$ is closest to
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A. $100g$
B. $150g$
C. $200g$
D. $50g$

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Last updated date: 13th Jun 2024
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Answer
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Hint:As we need to find out the solubility at $40^\circ C$, we can draw a line from the point $40^\circ C$ in the $x - axis$ to meet the curve. From this meeting point, we can draw a horizontal line to the $y - axis$ to find the solubility per $100g$. Dividing this value by two will give us our answer.

Complete step by step answer:
Since the temperature is given in the $x - axis$, we have to first draw a vertical line from the point $40^\circ C$ and extend it upwards till it meets the solubility curve. From the point of intersection of this line and the solubility curve, we should draw a horizontal line and extend it till it meets the $y - axis$. The point of intersection of this horizontal line and the $y - axis$ will give us the value on the $y - axis$ corresponding to the value on the $x - axis$. Since the intersection point is closest to $200g$, we can say that at a temperature of $40^\circ C$, the solubility of potassium nitrate ($KN{O_3}$) per $100g$ of water is $200g$. But we are required to find the quantity that dissolves in $50g$ of water at $40^\circ C$ , and not $100g$. Therefore, to get the answer per $50g$, we just have to divide the answer which we got per $100g$ by two. Hence,
Amount of potassium nitrate that dissolves in $50g$ of water at $40^\circ C$ $ = \dfrac{{200}}{2} = 100g$
Hence, the correct option to be marked is option A.

Note:
Solubility of a salt in water is dependent on lattice enthalpy (energy required to break the intermolecular forces of attraction between the salt molecules) and solvation enthalpy (energy released when the salt is dissolved in water). Thus, for a salt to be soluble in a solvent, its solvation enthalpy must be higher than its lattice enthalpy. Also note that as temperature increases, the solubility of most salts in a particular solvent tends to increase.