Question

The solubility of common salt at 30 C is 36g Calculate the amount of water required to prepare a saturated solution of 9g of salt at the given temperature

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Answer 1

Answer :-

Given :

• The solubility of common salt (NaCl) at 30°C is 36 g.

To find :

• Amount of water required to prepare a saturated solution of 90 g at the given temperature.

Solution :

We have:

⇒ Weight of NaCl = 90 g

⇒ Solubility of NaCl = 36 g

We know:

⇒ Solubility of solute:

• Weight of solute/Weight of solvent × 100

Hence:

⇒ 36 = 90/x × 100

⇒ 36 = 9000/x

⇒ 36x = 9000

⇒ x = 9000/36

∴ x = 250

∴,

The amount of water required to prepare the saturated solution is 250 g.

Answer 2

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■GIVEN THAT:-

▪︎A solubility of a common salt at 30°C is 36 g

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■ TO FIND:-

▪︎The amount of water required to prepare a saturated solution of 9 grams of salt at the given temperature.

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■ SOLUTION:-

We have,

◆ weight of NaCl=90 grams.

◆ solubility of NaCl= 36 grams

We know that;-

Solubility of solute:-

◆ weight of solute/weight of solvent ×100

36=90/x×100

36=9000/x

36x=9000

x=9000/36

x=250

♠︎ Therefore, the amount of water required to prepare a saturated solution is 250 grams.

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HOPE IT HELPS YOU DEAR!!!

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