The solubility of salt at $30^{\circ} C$ is 50. Calculate the weight (g) of water required to prepare a saturated solution containing 90 g of salt.

Open in App

Solution

Solubility (S) $=$ 50
Amount of salt (solute) $= w_{1} = 90 g$
Amount of water (solvent) $= w_{2} = ?$
Solubility (S) $= w_{1} / w_{2} \times 100$ ............. (4)
Weight of water (solvent) $= w_{2} = 100 w_{1} / S$
$= 100 \times 90 / 50 = 180 g$

Suggest Corrections

Similar questions

Q. Solubility of salt X in water is 32 at 30

^{o}

C. Calculate the amount of solvent present in 500 g of saturated solution of X.

Q. What weight of sodium nitrate will separate when a saturated solution containing 50 gram of water is cooled from

50^{\circ} C

30^{\circ} C

? The solubility of

N a N O_{3}

50^{\circ} C

and

30^{\circ} C

is 114 g and 86 g respectively.

(a) Find the solubility of $K N O_{3} a t 20^{\circ} C$ when the mass of the empty dish is 50 g, the mass of dish and solution is 65 g, while the mass of dish and residue is 54.3 g.

(b) What weight of sodium nitrate will separate when a saturated solution containing 50 gram of water is cooled from $50^{\circ} C t o 30^{\circ} C$ ? The solubility of $N a N O_{3} a t 50^{\circ} a n d 30^{\circ}$ is 114 g and 86 g respectively.

Q. To make a saturated solution, 136 g of a salt is dissolved in 500 g of water at 293 K. Find its solubility at this temperature.

Q. 50 g of a saturated aqueous solution of potassium chloride at

30^{o} C

is evaporated to dryness, when 13.2 g of dry KCl was obtained. Calculated the solubility of KCl in water at

30^{o} C

Join BYJU'S Learning Program

The solubility of salt at 30∘C is 50. Calculate the weight (g) of water required to prepare a saturated solution containing 90 g of salt.

Solubility (S) = 50Amount of salt (solute) =w1=90gAmount of water (solvent) =w2=?Solubility (S) =w1/w2×100 ............. (4)Weight of water (solvent) =w2=100w1/S=100×90/50=180g

The solubility of salt at $30^{\circ} C$ is 50. Calculate the weight (g) of water required to prepare a saturated solution containing 90 g of salt.

Solubility (S) $=$ 50
Amount of salt (solute) $= w_{1} = 90 g$
Amount of water (solvent) $= w_{2} = ?$
Solubility (S) $= w_{1} / w_{2} \times 100$ ............. (4)
Weight of water (solvent) $= w_{2} = 100 w_{1} / S$
$= 100 \times 90 / 50 = 180 g$