Solubility of $K N O_{3}$ at $333 K$ is $106 g$ . What mass of $K N O_{3}$ is needed to form a saturated solution in $75 g$ of water at $333 K$ ?
(Note: saturated solution is formed by dissolving substance in $100 g$ of water)

31 g

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70.7 g

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75 g

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79.5 g

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Solution

The correct option is D $79.5 g$
Mass of potassium nitrate present in $100 g$ of water in a saturated solution at $333 K$ is $106 g$ . Therefore, mass of potassium nitrate in $75 g$ of water in the saturated solution at $333 K$ will be:
$\frac{106}{100} \times 75 = 79.5 g$

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Q. Pragya tested the solubility of three different substances at different temperatures and collected the data as given below (results are given in the following table, as grams of substance dissolved in 100 grams of water to form a saturated solution).

a. What mass of potassium nitrate would be needed to produce a saturated solution of potassium nitrate in 50 grams of water at 313 K?

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Substance dissolved	Temperature in K
Substance dissolved	283	293	313	333	353
Potassium nitrate	21	32	62	106	167
Sodium chloride	36	36	36	37	37
Potassium chloride	35	35	40	46	54
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(a) What mass of potassium nitrate would be needed to produce a saturated solution of potassium nitrate in 50 grams of water at 313 K?

(b) Pragya makes a saturated solution of potassium chloride in water at 353 K and leaves the solution to cool at room temperature. What would she observe as the solution cools? Explain.

(c) Find the solubility of each salt at 293 K. What salt has the highest solubility at this temperature?

(d) What is the effect of change of temperature on the solubility of a salt?

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Solubility of KNO3 at 333 K is 106 g. What mass of KNO3 is needed to form a saturated solution in 75 g of water at 333 K? (Note: saturated solution is formed by dissolving substance in 100 g of water)

The correct option is D 79.5 gMass of potassium nitrate present in 100 g of water in a saturated solution at 333 K is 106 g. Therefore, mass of potassium nitrate in 75 g of water in the saturated solution at 333 K will be: 106100×75=79.5 g

Solubility of $K N O_{3}$ at $333 K$ is $106 g$ . What mass of $K N O_{3}$ is needed to form a saturated solution in $75 g$ of water at $333 K$ ?
(Note: saturated solution is formed by dissolving substance in $100 g$ of water)

The correct option is D $79.5 g$
Mass of potassium nitrate present in $100 g$ of water in a saturated solution at $333 K$ is $106 g$ . Therefore, mass of potassium nitrate in $75 g$ of water in the saturated solution at $333 K$ will be:
$\frac{106}{100} \times 75 = 79.5 g$